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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

• At t = 1 : ℑ (ρ)

• At t = 1 : ℑ (ρ) = 1 ⇒ u = ± √ 1 + v 2 and t = ± √ 1 + v 2 + iv ℜ (ρ) = −2uv = ∓v √ 1 + v 2 . To get descent for v → +∞ choose C3 using u = + √ 1 + v 2 , 0 ≤ v ≤ vmax ρ = it 2 �√ �2 � = i 1 + v2 + iv = i 1 + 2iv √ 1 + v2 � = i − 2v √ 1 + v2 • Connecting path C2: t = vmax (1 + i) + u, 0 ≤ u ≤ umax ≡ � 1 + v 2 max − vmax Evaluate integrals C2 : t2 = (vmax + u) 2 − v2 max + 2ivmax (vmax + u) � e C2 ixt2 � umax dt = e 0 i(vmax+u)2−v2 max −2xvmax(vmax+u) e du → 0 x → ∞ C1 : � C1 e ixt2 � ∞ dt = (1 + i) e 0 −2xu2 du = (1 + i) 1 � � π 1 π π = ei 4 2 2x 2 x This is the same result as the leading-order result obtained with method of stationary phase C3 : � C3 e ixt2 � ∞ dt = − 0 e ix−2xv√1+v2 � v √ 1 + v2 � + i dv where the minus sign reflects that C3 starts at infinity rather than 0. To use Watson’s lemma to get the full asymptotic expansion we want to have Instead of using v go back to t. We had therefore � C3 2v √ 1 + v 2 = s it 2 = i − 2v √ 1 + v 2 s = i − it 2 dt = e ixt2 dt = −e ix � ∞ ix i = −e 2 ix i = −e 2 0 t = (1 + is) 1 2 i 2 (1 + is) 1 2 e −xs ∞� � ∞ n=0 0 ds i 2 (1 + is) 1 2 ds e −xs (−is) n Γ � n + 1 2 ∞� (−i) n Γ � n + 1 � 2 n!Γ � � 1 2 n=0 66 � n!Γ � � ds 1 2 Γ (n + 1) x n+1

using the series expansion from the binomial theorem (22). Thus, we have Note: � ∞ 0 e ixt2 dt = 1 2 � π π ix i ei 4 − e x 2 ∞� (−i) n Γ � n + 1 � 2 Γ � 1 � 1 x 2 n+1 • since the integrand is analytic everywhere the return contour can also be deformed away from the steepest-descent contout C3 without affecting the value of the integral. This can be exploited to make the integration simpler. • The integral over the deformed contour has to be amenable to an asymptotic method like Laplace’s method. Deform the steepest-descent contour C3 into a contour C4 that simplifies the calculation C4 : t = 1 + iv, 0 ≤ v ≤ vm with an arbitrary contour from tm = 1 + ivm to contour C3. � C3 e ixt2 � dt = C4 e ixt2 dt ∼ i � ǫ 0 n=0 e ix(1+iv)2 dv = i � ǫ 0 e ix(1−v2 )−2xv dv since for v > ǫ the contributions to the integral are exponentially small it is sufficient to integrate only over the interval [0, ǫ]. Using Laplace’s method and extending the integration again to ∞ one gets (cf. treatment of higher order-terms in φ in Sec. 2.3.3) � C4 e ixt2 dt = ie ix Compare with the result using C3: (2n!) 1 = n!22n n! 2n 2 since Γ (x) = (x − 1)Γ(x − 1). = ie ix = ie ix 2n − 1 2n − 2 2 2 ∞� � ǫ n=0 ∞� n=0 ∞� n=0 0 1 n! . . . 2 1 2 2 1 n! (−ix)n v 2n e −2xv dv (−ix) n (2x) 2n+1 (−i) n (2n!) n! 2 2n+1 Thus, both contours give exactly the same result. Note: � ∞ 0 1 xn+1 2n − 1 2n − 3 = 2 2 u 2n e −u du x → ∞ . . . 3 1 2 2 = Γ � n + 1 � 2 Γ � � 1 • the path of steepest descend avoids the oscillations in the integral and therefore allows Laplace’s method; but any other path that allows Laplace’s method works as well. A path that is tangential to the contour with constant ψ will have only slow oscillations near the saddle, which can be captured as higher-order corrections. 67 2

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