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# Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

## analogously for C2. �

analogously for C2. � C1 ∼ � +ǫ 0 ≤ e −4x � −ǫ 1ds e −4x(1+s)2 e ix(−(−1−s)3 +5(−1−s)) ds 0 ≤ O(e −4x ) C3 : use contour that is tangential to the steepest descent contour t = i + s − ǫ ≤ s ≤ ǫ v = 1 u = s � C3 ∼ � +ǫ −ǫ ∼ 1 √ x e −2x ∼ 1 √ x e −2x ∼ � π x e−2x e x{−s2 −2} e ix{−s 3 } ds � + √ xǫ − √ xǫ � +∞ −∞ e −u2 − √ i u e x 3 du e −u2 � 1 − i √ u x 3 � du Thus the integrals over contours C1,2 are exponentially smaller than that over contour C3 I(x) ∼ 1 � π 2 x e−2x 2.5.2 Complex x and the Stokes Phenomenon Generalize the integral further to allow x to be complex • now the asymptotic expansion can depend on the argument of x depending on the argument of x different terms can be dominant and subdominant • the interchange of dominance and subdominance when the argument of the expansion variable x is varied is called Stokes Phenomenon Example: Again I(x) = � 1 0 e−4xt2 cos (5xt − xt 3 ) dt for large complex x = X + iY The saddle points do not depend on x ⇒ dominant contributions still will come from the endpoints at t = ±1, which are maxima, or from the saddle point, t = i. Need to consider steepest descent contours near the saddle points: t = −1 + U + iV with |U|, |V | ≪ 1: ρ = −V 3 + 3 (−1 + U) 2 V − 4 (−1 + U) 2 + 4V 2 − 5V + +i � − (−1 + U) 3 + 3 (−1 + U) V 2 − 8 (−1 + U) V + 5 (−1 + U) � = −4 + 3V + 8U − 5V + h.o.t. + i {1 − 5 − 3U + 5U + 8V } = −4 + 8U − 2V + i {−4 + 2U + 8V } + h.o.t. 72

Write xρ = (X + iY ) ρ ≡ Φ + iΨ We need contour with Ψ = const. to leading non-trivial order in U and V , Ψ = X {−4 + 2U + 8V } + Y {−4 + 8U − 2V } = −4 (X + Y ) � �� � value of ψ at endpoint +U (2X + 8Y ) + V (8X − 2Y ) Thus, X + 4Y V = − U + h.o.t. 4X − Y Then to leading non-trivial order in U and V and Φ = X (−4 + 8U − 2V ) − Y (−4 + 2U + 8V ) = 4 (Y − X) + U (8X − 2Y ) + V (−2X − 8Y ) � � (X + 4Y )2 = 4 (Y − X) + U 8X − 2Y + 2 4X − Y = 4 (Y − X) + U 32X2 − 8XY − 8XY + 2Y 2 + 2X 2 + 16XY + 32Y 2 4X − Y Φ = 4 (Y − X) + U 34 (X2 + Y 2 ) 4X − Y � X + 4Y dt = 1 − i 4X − Y � dU and we get for the contribution from the end point t = −1 � C1 = 1 � X + 4Y 1 − i 2 4X − Y = − 1 � X + 4Y 1 − i 2 4X − Y � � −ǫ = − 1 4X − Y − iX − 4iY e−4x(1+i) 2 34 (X + iY ) (X − iY ) = − 1 1 e−4x(1+i) 2 34x = 1 − 4 e−4x(1+i)i 2 34x e 4(Y −X) e −4i(X+Y )+O(U2 34(X ) e 2 +Y 2 ) U 4X−Y dU 0 � e 4(Y −X) e −4i(X+Y ) 4X − Y 34 (X2 + Y 2 ) � 4 − C3 Y + iX X − iY Compare this contribution with that from � ∼ 1 � π 2 x e−2x They have the same dominant exponential for � 4ℜ ((X + iY ) (1 + i)) = 2X 73

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