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Lecture Note Sketches Hermann Riecke - ESAM Home Page

• 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 **Note**s: � � 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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Lecture Note Sketches Perturbation

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4 Fronts and Their Interaction 87 4

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References [1] P. Coullet, C. Elphi

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1.1.1 The Mathieu Equation Consider

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• case δ1 = + 1 2 � � 1 ü2

- Page 11 and 12: • we can assume from the start th
- Page 13 and 14: 1.1.2 Floquet Theory In the discuss
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- Page 17 and 18: Goal: determine α(δ, ǫ). For a r
- Page 19 and 20: Forcing Strength ε 1 0.5 0 -0.5 n=
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- Page 23 and 24: O(ǫ): with Note: y0(ψ, T) = R(T)
- Page 25 and 26: i.e. Now the nonlinear problem: Fix
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- Page 31 and 32: • using the other left 0-eigenvec
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- Page 39 and 40: with O(1) forcing (cf. in Sec.1.2.1
- Page 41 and 42: Thus, if ∆ > 0 both solutions R2
- Page 43 and 44: Here we can get an integral express
- Page 45 and 46: which diverge at the upper limit fo
- Page 47 and 48: as in the case a > 0. But: now the
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- Page 51 and 52: First consider maximum at the lower
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- Page 57 and 58: It looks like a case for Watson’s
- Page 59 and 60: Extend integration to (−∞, +∞
- Page 61: Bound the integral term ���
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- Page 67 and 68: using the series expansion from the
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- Page 95 and 96: Analogously for x > xm: 0 = ǫ∂Tx
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