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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Why is this method called ‘Steepest Descent’? ρ(t) = φ(t) + iψ(t) is an analytic function of t ≡ u + iv. It therefore satisfies the Cauchy- Riemann conditions ∂φ ∂ψ ∂φ = = −∂ψ ∂u ∂v ∂v ∂u Consider (u, v) as a two-dimensional vector. One can then write � � ∂f ∂f , = ∇f ∂u ∂v One then gets ∇φ · ∇ψ = ∂φ ∂ψ ∂u ∂u ∂φ ∂ψ + ∂v ∂v Lines of constant ψ are orthogonal to ∇ψ ∇φ is orthogonal to ∇ψ ⎫ ⎬ ⎭ ∂φ ∂ψ = ∂u ∂u + � − ∂ψ � ∂φ ∂u ∂u = 0 ⇒ ψ = const. � ∇φ Since ∇φ gives the direction of steepest ascent, going away from a maximum along lines ψ = const. amounts to going in the direction of steepest descent. 2.5.1 Saddle Points So far the maxima were at the end of the contours. Consider now local maxima of φ in the interior of the integration interval. For analytic ρ = φ + iψ the functions φ and ψ are harmonic ∆φ = 0 ∆ψ = 0 They cannot take on maxima or minima in the interior of a bounded domain. At the maxima of φ along the lines of constant ψ one has dψ ds = 0 and dφ ds = 0 ⇒ dρ ds = 0 ⇒ ρ′ = 0 By Cauchy-Riemann all directional derivatives of ψ and φ vanish ⇒ the maxima of φ along the lines of constant ψ are saddles in the complex plane. At a saddle multiple lines of constant ψ intersect. Example: Saddle point of e xt2 ρ(t) = t 2 = (u + iv) 2 = u 2 − v 2 + 2iuv and ρ ′ (t) = 2t = 2u + i2v The only saddle point is The steepest paths are given by uv = 0: u = 0 ρ = −v 2 v = 0 ρ = +u 2 ρ ′ = 0 u = 0 = v 68 steepest descent steepest ascent

Example: Behavior near the saddle point of e x(sinht−t) near t = 0 At t = 0 we have ρ = sinh t − t = 0 ρ ′ = cosh t − 1 = 0 ρ ′′ = sinh t = 0 ρ ′′′ = cosh t = 1 Thus, also the second derivative vanishes, but not third one (third-order saddle point). Steepest paths ρ = sinh (u + iv) − u − iv = sinh u cosv + i cosh u sinv − u − iv cosh u sinv − v = 0 ⇔ v = 0 or u = cosh −1 � v � sin v The second condition defines two lines because cosh −1 is double-valued. Up Down Up π v Up Figure 11: Saddle of e x(sinht−t) . Down u Down Example: I(x) = � 1 0 e−4xt2 cos (5xt − xt 3 )dt for large x This is not a Laplace integral because x appears also in the cosine. i) Try nevertheless to argue that integral dominated by small values of t because of the exponential. One would get � ∞ I(x) ∼ 0 e −4xt2 dt = 1 � π 2 4x wrong But: exponential has decayed only for t > x −1 2 ⇒ xt ∼ x 1 2 → ∞ and argument of cos is large and the rapid oscillations of the cos lead to substantial cancellation. ii) For t = O(x−1 2) the second term in the cos is small: xt3 = O(x−1 2) ⇒ tempting to ignore that term. Then � 1 I(x) ∼ e −4xt2 � ∞ cos (5xt) dt ∼ e −4xt2 cos (5xt) dt = = 1 2 = 1 2 0 0 � +∞ e −∞ −4xt2 � +∞ +5ixt 1 dt = e 2 −∞ −4x(t−5 8i)2 − 25 16x dt = � π 4x e−25 16x wrong 69

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Lecture Note Sketches Perturbation

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4 Fronts and Their Interaction 87 4

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References [1] P. Coullet, C. Elphi

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1.1.1 The Mathieu Equation Consider

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• case δ1 = + 1 2 � � 1 ü2

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• we can assume from the start th

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1.1.2 Floquet Theory In the discuss

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• 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
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- 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
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- 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 ���
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- Page 65 and 66: using � ∞ 0 e−u ln u du = γ
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- Page 73 and 74: Write xρ = (X + iY ) ρ ≡ Φ + i
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- Page 95 and 96: Analogously for x > xm: 0 = ǫ∂Tx
- Page 97 and 98: - L = 0 corresponds to a pure gas p