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• since the remainder

• since the remainder is O(x−3 2) it is clear that it could not be obtained by integration by parts. 2.4.1 Method of Stationary Phase As before, if ψ ′ (t) = 0 somewhere in the interval [a, b] integration by parts may not work ⇒ use the method of stationary phase: • Near the stationary point the oscillations of e ixψ(t) are less rapid leading to less cancellation ⇒ the dominant contributions to the integral come from the stationary point. ⇒ we can restrict the integral to the vicinity of the stationary points Show the dominance of the contributions from the stationary point: One can always divide the integration domain into sections such that the stationary point ends up at an endpoint a of a subintegral. It is therefore sufficient to consider c ≡ a, ψ ′ (c) = 0, ψ ′ (t) �= 0 for t ∈ (a, b] Note: � a+ǫ I(x) = f(t)e ixψ(t) � b dt + • the second integral is of O(x −1 ) since ψ ′ (t) �= 0 (cf. (28)) a a+ǫ f(t)e ixψ(t) dt To get the leading behavior of the first integral expand f(t) = f(a) + . . ., and ψ(t) = ψ(a) + 1 p! ψ(p) (a) (t − a) p where ψ (p) (a) is the first non-vanishing derivative of ψ(t) at t = a, � a+ǫ I(x) ∼ a 1 ix[ψ(a)+ [f(a) + . . .] e p! ψ(p) (a)(t−a) p ] dt Extending the integral to ∞ introduces errors of O( 1) since there are no stationary points x in the added interval (Riemann-Lebesgue) I(x) ∼ f(a)e ixψ(a) � ∞ 0 1 ix e p! ψ(p) (a) (t−a) p dt As before, the integral can be obtained by deforming the contour to go out to infinity along π ±i a ray e 2p (cf. (29)): We want (t − a) p p! = −u ixψ (p) (a) 62

π i to ensure decay of the exponential along that ray need to choose e 2p for ψ (p) π i > 0 and e −2p for ψ (p) < 0 ψ (p) � π i p! (a) > 0 : t − a = e 2p ψ (p) π −i (a) < 0 : t − a = e 2p xψ (p) (a) � 1 p u 1 p � − p! xψ (p) (a) � 1 p u 1 p I(x) ∼ f(a)e ixψ(a) � π ±i e 2p ± p! xψ (p) � 1 p (a) � ∞ 1 1 u p 0 p −1 e −u du Thus: the method of stationary phase yields the leading-order term of I(x) as � I(x) ∼ f(a)e ixψ(a) � π ±i e 2p ± p! xψ (p) � 1 p Γ (a) � 1 p p x → ∞ (30) where the sign in ± has to agree with that of ψ (p) (a) and the endpoint a is the point of stationary phase. Example: I(x) = � ∞ 0 cos (xt2 − t)dt To use the method stationary phase rewrite the integral �� ∞ � I(x) = ℜ and identify f(t) = e −it Stationary point is at t = 0 with ψ ′′ (0) = 2 > 0 Notes: � � π i π I(x) ∼ ℜ e 4 x 1 − • (30) gives the leading-order behavior x p 0 e i(xt2 −t) dt ψ(t) = t 2 � 1 = 2 1 � π 2 2x x → ∞ • Terms omitted include terms O( 1) arising from the integration away from the station- x ary point ⇒ to obtain higher-order approximations is not so easy, since contributions may arise from the whole integration interval [a, b]. This is to be compared with the Laplace’s method, where the contributions from the domain away from the maximum are exponentially rather than algebraically small ⇒ use method of steepest descend (Sec.2.5). • If f(a) = 0 it is not clear whether the integral is still dominated by the stationary point 63

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