- Text
- Solution,
- Integral,
- Equation,
- Forcing,
- Amplitude,
- Term,
- Integration,
- Solutions,
- Consider,
- Maximum,
- Lecture,
- Note,
- Sketches,
- Hermann,
- Riecke,
- Home,
- People.esam.northwestern.edu

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Integral now exponentiall small, but still wrong (can be checked by expanding the cos and comparing the order of the omitted term with the retained term. ) iii) Use method of steepest descent. Rewrite with I(x) = 1 2 � +1 −1 e −4xt2 +5ixt−ixt3 dt = 1 2 ρ(t) = −it 3 − 4t 2 + 5it Phases at the end points differ from each other ℑ (ρ(t = ±1) = ±4 Identify contours C1,2 of constant ℑ (ρ) using t = u + iv **Note**: � +1 −1 e xρ(t) dt ρ = −i � u 3 + 3iu 2 v − 3uv 2 − iv 3� − 4 � u 2 − v 2 + 2iuv � + 5iu − 5v = −v 3 + 3u 2 v − 4u 2 + 4v 2 − 5v � �� � φ • φ is even in u and ψ is odd in u. ψ = 4σ at t = σ with σ = ±1: 3uv 2 − 8uv + 5u − u 3 − 4σ = 0 +i � −u 3 + 3uv 2 − 8uv + 5u � � �� � ψ v = 1 � 8u ± 6u � 64u2 − 12u (5u − u3 � − 4σ) = 1 � 4u ± 3u √ u2 + 3u4 � + 12σu We need the contours passing through t = σ, i.e. u = σ and v = 0 σ = +1 : v2 = 1 � 4u − 3u √ u2 + 3u4 � + 12u σ = −1 : v1 = 1 � 4u + 3u √ u2 + 3u4 � − 12u Saddles Up Down v Down Up Down Up Down Up Down Down Up Figure 12: Contours with ψ = const and saddle points for e−4xt2 +5ixt−ixt3 . 70 u

Limiting behavior of the contours v ± 1,2 σ = +1 : v2 → − u √ 3 σ = −1 : v1 → + u √ 3 for u → +∞ v2 → − 2 1 √ √ 3 u for u → 0 + for u → −∞ v1 → − 2 1 √ √ 3 −u for u → 0 − Limiting behavior of real part φ along the contours v ± 1,2 for u → ±∞ and v → −∞ Use u 2 → 3v Thus σ = ±1 : φ → −v 3 + 3u 2 v → −v 3 + 9v 3 = 8v 3 → −∞ for v → −∞ • If we deformed the integration contour to include the singularity at u = 0 the main contribution to the integral would arise at the singularity. Poor choice • Integrate from t = ±1 outward to u → ±∞ – the main contribution to this contour integral arises near t = ±1 – we need to connect the end points with an additional contour C3 Identify all saddle points of ρ(t) dρ dt = −3it2 − 8t + 5i = 0 � t (s) 5 1 = 3i t (s) 2 = i t (s) 1,2 = − 8 ± √ 64 − 60 6i Along the contours with ψ = const. the real part φ can only have a local maximum at one of the two saddle points ⇒ φ does not have a local maximum along the contours v1,2 Identify contours with ψ = const. through the saddle point t = i ψ(t = i) = 0 ⇒ u = 0 or 3v 2 − 8v + 5 − u 2 = 0 ⇒ v3,4 = 4 ± √ 1 + 3u 2 The contour v4 is asymptotic to the contours v1,2. The integral along C3 does not vanish: φ reaches a maximum at the saddle at t = i. Therefore the dominant contribution to that integral arises near that saddle. � 2I(x) ∼ e xρ � dt + e xρ � dt + e xρ dt ∼ C1 C3 � t=−1−O(ǫ) e xρ � t=i+O(ǫ) dt + e xρ � t=1 dt + t=−1 C2 t=i−O(ǫ) t=1+O(ǫ) e xρ dt We can deform the contours slightly away from the contour ψ = const. to simplify the integrals C1 : t = −1 − s 0 ≤ s ≤ ǫ 71 3

- 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 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 (−∞, +∞
- 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: Example: Behavior near the saddle p
- 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