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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Integral now

Integral now exponentiall small, but still wrong (can be checked by expanding the cos and comparing the order of the omitted term with the retained term. ) iii) Use method of steepest descent. Rewrite with I(x) = 1 2 � +1 −1 e −4xt2 +5ixt−ixt3 dt = 1 2 ρ(t) = −it 3 − 4t 2 + 5it Phases at the end points differ from each other ℑ (ρ(t = ±1) = ±4 Identify contours C1,2 of constant ℑ (ρ) using t = u + iv Note: � +1 −1 e xρ(t) dt ρ = −i � u 3 + 3iu 2 v − 3uv 2 − iv 3� − 4 � u 2 − v 2 + 2iuv � + 5iu − 5v = −v 3 + 3u 2 v − 4u 2 + 4v 2 − 5v � �� � φ • φ is even in u and ψ is odd in u. ψ = 4σ at t = σ with σ = ±1: 3uv 2 − 8uv + 5u − u 3 − 4σ = 0 +i � −u 3 + 3uv 2 − 8uv + 5u � � �� � ψ v = 1 � 8u ± 6u � 64u2 − 12u (5u − u3 � − 4σ) = 1 � 4u ± 3u √ u2 + 3u4 � + 12σu We need the contours passing through t = σ, i.e. u = σ and v = 0 σ = +1 : v2 = 1 � 4u − 3u √ u2 + 3u4 � + 12u σ = −1 : v1 = 1 � 4u + 3u √ u2 + 3u4 � − 12u Saddles Up Down v Down Up Down Up Down Up Down Down Up Figure 12: Contours with ψ = const and saddle points for e−4xt2 +5ixt−ixt3 . 70 u

Limiting behavior of the contours v ± 1,2 σ = +1 : v2 → − u √ 3 σ = −1 : v1 → + u √ 3 for u → +∞ v2 → − 2 1 √ √ 3 u for u → 0 + for u → −∞ v1 → − 2 1 √ √ 3 −u for u → 0 − Limiting behavior of real part φ along the contours v ± 1,2 for u → ±∞ and v → −∞ Use u 2 → 3v Thus σ = ±1 : φ → −v 3 + 3u 2 v → −v 3 + 9v 3 = 8v 3 → −∞ for v → −∞ • If we deformed the integration contour to include the singularity at u = 0 the main contribution to the integral would arise at the singularity. Poor choice • Integrate from t = ±1 outward to u → ±∞ – the main contribution to this contour integral arises near t = ±1 – we need to connect the end points with an additional contour C3 Identify all saddle points of ρ(t) dρ dt = −3it2 − 8t + 5i = 0 � t (s) 5 1 = 3i t (s) 2 = i t (s) 1,2 = − 8 ± √ 64 − 60 6i Along the contours with ψ = const. the real part φ can only have a local maximum at one of the two saddle points ⇒ φ does not have a local maximum along the contours v1,2 Identify contours with ψ = const. through the saddle point t = i ψ(t = i) = 0 ⇒ u = 0 or 3v 2 − 8v + 5 − u 2 = 0 ⇒ v3,4 = 4 ± √ 1 + 3u 2 The contour v4 is asymptotic to the contours v1,2. The integral along C3 does not vanish: φ reaches a maximum at the saddle at t = i. Therefore the dominant contribution to that integral arises near that saddle. � 2I(x) ∼ e xρ � dt + e xρ � dt + e xρ dt ∼ C1 C3 � t=−1−O(ǫ) e xρ � t=i+O(ǫ) dt + e xρ � t=1 dt + t=−1 C2 t=i−O(ǫ) t=1+O(ǫ) e xρ dt We can deform the contours slightly away from the contour ψ = const. to simplify the integrals C1 : t = −1 − s 0 ≤ s ≤ ǫ 71 3

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