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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

2.5 Method of Steepest

2.5 Method of Steepest Descent Consider now Laplace integrals with fully complex exponent with h(t) and ρ(t) analytic functions. � b I(x) = h(t)e xρ(t) dt a Approach: deform the integration contour such that the imaginary part ψ of ρ ≡ φ + iψ is constant I(x) = e ixψ � h(t)e xφ(t) dt For the resulting integral one can then use Laplace’s method. Note: C • one could use also contours of constant real part φ and then use the method of stationary phase for the resulting integral. However, only the leading-order contribution is given by the vicinity of the point of stationary phase, while with Laplace’s method the whole asymptotic expansion is determined by the neighborhood of the maximum. Example: Compute the full asymptotic series for I(x) = � 1 0 ln t eixt dt Integration by parts does not work since ln t diverges at t = 0 Method of stationary phase does not work because there is no stationary point. Deform the integration path to make the resulting integrals suitable for Laplace’s method: need to keep the imaginary part of ixt constant: e xφ(t) Use three contours ℑ (ixt) = const ⇔ ℜ (t) = const C1 : t = is, 0 ≤ s < T C2: t = iT + s, 0 ≤ s ≤ 1, C3: t = 1 + is, T ≥ s ≥ 0 and let T → ∞ � I(x) = i ln (is)e −xs � ds + ln (iT + s)e −xT+ixs � ds + i ln (1 + is) e ix−xs ds C1 0 C2 i) Integral � C2 : it vanishes for T → ∞ because of the factor e−xT . ii) Integral � C1 : exponential has maximum at s = 0, but the ln(is) cannot be expanded at s = 0 but the integral can be done exactly � ∞ i ln (is)e −xs ds = i � ∞ (ln i + ln u − ln x) e x −u du = − 1 x 0 64 � π + iγ 2 � ln x − i x C3

using � ∞ 0 e−u ln u du = γ ≈ 0.572 . . . (Euler’s constant). iii) Integral � C3 : use Watson’s lemma with global maximum at s = 0, α = 0, β = 1, and using ∞� (−is) ln (1 + is) = − n n get ie ix � C3 ln (1 + is) e −xs ds = ie ix ∞� n=1 n=1 Combined we get I(x) = − 1 � π � ∞� ln x + iγ − i + ieix x 2 x Notes: (−1)(−i) n Γ(n + 1) x n+1 n n=1 (−i) n (n − 1)! x n+1 x → ∞ x → ∞ • in this example ρ = it ⇒ ℑ(ρ(t = 0)) and ℑ (ρ(t = 1)) differ from each other ⇒ there is no constant-phase contour that connects ρ(t = 0) and ρ(t = 1) ⇒ we needed three contours, to be chosen such that two lead to Laplace integrlas and one gives a vanishing contribution. Example: Determine the full asymptotic behavior of I(x) = � 1 0 eixt2 dt To get the leading-order behavior the method of stationary phase is sufficient: ψ = t 2 with stationary point t = 0 and ψ ′′ (a) = 2 > 0. I(x) ∼ 1 π ei 4 2 � 2π x 2 To get the full asymptotic behavior is difficult with this method. Use steepest descent instead. Deform contour into contours along which the imaginary part of ρ ≡ it 2 is constant. Note: • t is to be considered a complex variable t ≡ u + iv: ℑ (ρ) = u 2 − v 2 Identify suitable contours: • At t = 0: ℑ (ρ) = 0 ⇒ u = ±v u = −v: u = +v: e xρ = e ixt2 = e ix(u−iu)2 = e +2xv2 ℜ (ρ) = −2uv = ∓2v 2 e ixt2 = e ix(v+iv)2 = e −2xv2 ascent, diverges for u → ∞ descent ⇒ to get steepest descent path choose C1 : t = v + iv, 0 ≤ v ≤ vmax with vmax → ∞ eventually 65

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