- 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

Consider the two linearly independent solutions � u1(0) = 1 u1(t) with ˙u1(0) = 0 u2(t) with � u2(0) = 0 ˙u2(0) = 1 Since u1,2 are linearly independent their Wronskian is non-zero � � W(u1(t), u2(t)) = � � u1(t) � u2(t) � � ˙u1(t) ˙u2(t) � = � � � � u1(0) � u2(0) � � ˙u1(0) ˙u2(0) � = 1 �= 0 Use u1,2 to express the general solution as u(t) = c1u1(t) + c2u2(t) where the c1,2 are given by the initial conditions c1 = u(0) c2 = ˙u(0) (2) Now consider u(t + T). For this discrete time translation u1,2(t + T) are also solutions of (1) and can be expanded in terms of u1,2(t) using (2). This gives uj(T + t) = uj(T)u1(t) + ˙uj(T)u2(t) j = 1, 2 u(t + T) = c1u1(t + T) + c2u2(t + T) The relation u(t + T) = ρu(t) yields then = c1 [u1(T)u1(t) + ˙u1(T)u2(t)] + c2 [u2(T)u1(t) + ˙u2(T)u2(t)] c1 [u1(T)u1(t) + ˙u1(T)u2(t)] + c2 [u2(T)u1(t) + ˙u2(T)u2(t)] = ρ [c1u1(t) + c2u2(t)] Since u1(t) and u2(t) are linearly independent we get c1u1(T) + c2u2(T) = ρc1 c1 ˙u1(T) + c2 ˙u2(T) = ρc2 requiring � ��� u1(T) − ρ u2(T) ˙u1(T) ˙u2(T) − ρ **Note**: ρ 2 − ρ (u1(T) + ˙u2(T)) � �� � ≡2K • ρ is called the Floquet multipler � � � � = 0 + u1(T) ˙u2(T) − ˙u1(T)u2(T) = 0 � �� � W=1 ρ 2 − 2Kρ + 1 = 0 ρ1,2 = K ± √ K 2 − 1 14

• it is also convenient to introduce α via ρ1 = e iαT which characterizes the Floquet exponent iαT. Because W = 1 one has ρ1ρ2 = 1 implying ρ2 = e −iαT . The evolution of u(t) during one period of Q(t) is therefore determined by K 1. K = 1: ρ1,2 = 1 if K = 1 because ρ = 1 one has a T-periodic solution. 2. K = −1: ρ1,2 = −1 if K = −1 because ρ = −1 one has a 2T-periodic solution u(t + 2T) = −u(t + T) = u(T) 3. |K| < 1: ρ1,2 are complex with |ρ1,2| = 1, α ∈ R 4. |K| > 1: |ρ1| > 1 and |ρ2| < 1 are real and α ∈ iR For |K| �= 1 one obtains two solution u (1,2) (t) satisfying Factoring out the Floquet multiplier e iαT via we get u (1) (t + T) = ρ1u (t) (t) = e iαT u (1) (t) u (2) (t + T) = ρ2u (2) (t) = e −iαT u (2) (t) u (1) (t) = e iαt U (1) (t) U (1) (t + T) = u (1) (t + T)e −iα(t+T) = e iαT u (1) (t)e −iα(t+T) = U (1) (t) i.e. U (1) (t) is T-periodic. Analogously U (1) (t) and U (2) (t) are T-periodic. u (2) (t) = e −iαt U (2) (t) The u (1,2) (t) are independent linear combinations of the two linearly independent solutions u1(t) and u2(t) ⇒ u (1,2) (t) are also linearly independent. More formally: assume there are k1 �= 0 �= k2 with k1u (1) (t) + k2u (2) (t) = 0 for all t k1e iαt U (1) (t) + k2e −iαt U (2) (t) = 0 k1e iα(t+T) U (1) (t + T) + k2e −iα(t+T) U (2) (t + T) = 0 Using U (1,2) (t + T) = U (1.2) (t) non-zero ki requires e iαt U (1) (t)e −iα(t+T) U (2) (t) − e −iαt U (2) (t)e iα(t+T) U (1) (t) = 0 15

- 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: 1.1.2 Floquet Theory In the discuss
- 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 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