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## Why is this method

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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