Notes 14 The Steepest-Descent Method

Transcription

1 ECE 638 Fall 17 David R. Jackson Notes 14 The Steepest-Descent Method Notes are adapted from ECE

2 Steepest-Descent Method Complex Integral: Ω g z I f z e dz Ω = C The method was published by Peter Debye in 199. Debye noted in his work that the method was developed in a unpublished note by Bernhard Riemann (1863). Peter Joseph William Debye (March 4, 1884 November, 1966) was a Dutch physicist and physical chemist, and Nobel laureate in Chemistry. Georg Friedrich Bernhard Riemann (September 17, 186 July, 1866) was an influential German mathematician who made lasting contributions to analysis and differential geometry, some of them enabling the later development of general relativity.

3 Complex Integral: Ω g z I f z e dz Ω = C We want to obtain an approximate evaluation of the integral when the real parameter Ω is very large. The functions f (z) and g(z) are analytic (except for poles or branch points), so that the path C may be deformed if necessary (possibly adding residue contributions or branch-cut integrals). Saddle Point (SP): g ( z ) = (, ) g( xy, ) g xy x =, = y 3

4 Path deformation: If the path does not go through a saddle point, we assume that it can be deformed to do so. If any singularities are encountered during the path deformation, they must be accounted for (e.g., residue of captured poles). y C = + z C C' C p z p X C p = π Res ( p ) F z dz i F z C p x C 4

5 Denote g( z) = u( z) + iv( z) Cauchy Reimann eqs.: Hence u v = x y u = y v x u v v u = = = x x y y x y y u = y 5

6 or u x u y + = If u xx <, then u yy > Near the saddle point: 1 u( x, y) u( x, y) + uxx ( x x) 1 + uyy y y + u x x y y xy Ignore (rotate coordinates to eliminate) 6

7 1 1 u x y u x y u x x u y y u x x y y (, ) (, ) + ( ) + ( ) + ( )( ) xx yy xy In the rotated coordinate system: 1 1 u x y u x y u x x u y y (, ) (, ) + ( ) + ( ) xx yy Assume that the coordinate system is rotated so that u < u > xx yy Note: The angle of rotation necessary to do this will be clear later (it is the departure angle θ of the steepest-descent path). 7

8 The u (x, y ) function has a saddle shape near the SP: ux (, y ) y z x 8

9 uxy (, ) y z x y x Note: The saddle does not necessarily open along one of the principal axes (only when u xy (x, y ) = ). 9

10 A descending path is one on which the u function decreases away from the saddle point. uxy (, ) y x C z 3D view of descending path 1

11 Along any descending path C we will have convergence: ( Ω ) = Ωg z I f z e dz = = C Ω g z g( z ) Ω g z e f z e dz C iω vz vz Ω uz uz Ω g z e f z e e dz C Exponentially decreasing function 11

12 g z iω vz vz Ω uz uz Ω I e f z e e dz Ω = Behavior on a descending path C Let e s u z u z ( u( z ) u z ) s Ω Ω = e e Ωs 1 Note: The parameter s is a measure of distance along the path from the saddle point. (Negative s means before we get to the saddle point, positive s means after we leave the saddle point.) When Ω is large most of contribution is from near z. 3 Ω=1 s 1

13 Sketches of a descending path: uxy (, ) s u z u z y y x C C z z s < s = s > x e Ωs 1 3 Ω=1 s 13

14 Along any descending path: g( z ) Ω g z I e f z e dz Ω = ~ = Ω g( z) g( z ) Ω g z f z e e dz iω vz vz Ω uz uz Ω g z f z e e e dz C C C Ω g z Both the phase and amplitude change along an arbitrary descending path C. Important Point: If we can find a path along which the phase does not change (v is constant), the integrand will have a purely exponentially decaying behavior (no phase term), making the integral easy to evaluate. 14

15 Choose path of constant phase: C : v z = v z = constant y C z s < s = s > x 15

16 Gradient Property (proof on next slide): uxy, C is parallel to Hence C is either a path of steepest descent () or a path of steepest ascent (SAP). : u(x,y) decreases as fast as possible along the path away from the saddle point. SAP: u(x,y) increases as fast as possible along the path away from the saddle point. 16

17 Proof u u = xˆ + yˆ x v v = xˆ + yˆ x u y v y u v u v u v= + x x y y u u u u = + x y y x = C.R. Equations Hence, u v C y v Also, v C v ( is constant on C ) u x Hence u C 17

18 iω vz vz Ω uz uz Ω g z I Ω ~ f z e e e dz C Because the v function is constant along the, we have: or Ω u( z) u( z ) I ~ f z e e dz Ω Ω g z Ω g( z) g( z ) I Ω ~ f z e e dz Ω g z Real 18

19 Local behavior near SP 1 g z g z g z z z g z z z + ( ) + ( ) so 1 g z g z g z z z ( ) y z +dz Denote g z = Re iα z z = dz = re iθ z r θ x 1 ( + ) i g z g z Rr e α θ 19

20 1 g z g z Rr e α θ i( ) + u z 1 u z Rr v z 1 v z Rr cos( α + θ) sin ( α + θ) SAP: n α + θ = + πn α α =,1: θ =, + π 1 uz uz ( ) = Rr vz vz = : α + θ = π + πn α π α π n = 1, : θ =, + 1 uz uz ( ) = Rr vz vz = Note: The two paths are 9 o apart at the saddle point.

21 y SAP SP u 9 o u increases u u decreases x Note: v(x,y) is constant along both paths. 1

22 The landscape of the function u(z) near z y SAP Hill z uz = uz Valley C Valley Hill x uz = uz

23 uxy (, ) y x z SAP 3

24 Ω Ω u( z) u( z ) I f z e e dz ~ Ωg z Set s u( z ) u( z) = This defines z= zs () uxy (, ) s y x s > s < z z 4

25 We then have Ωg ( z + ) Ωs dz Ω ~ I f z e e ds ds or ~ dz Ω s Ω g z Ω ds s= I f z e e ds + (leading term of the asymptotic expansion) Hence ~ I f z e Ωg z dz ds Ω s= π Ω 5

26 To evaluate the derivative: = ( ) g( z ) s u z u z = g z (Recall: v is constant along.) ds s = g z dz At the SP this gives =. Take one more derivative: ds ds d s s = g z dz dz dz 6

27 At s = : ( z ) ds g = dz Z 1/ so dz = ds g z s= 1/ Hence, we have 1/ Ωg( z ) π I( Ω) ~ f ( z ) e g ( z ) Ω 7

28 There is an ambiguity in sign for the square root: dz = ds g z s= 1/ y To avoid this ambiguity, define z θ arg dz ds s= = θ x 8

29 The derivative term is therefore dz = ds g z s= iθ e Hence π I Ω f z e e Ω g z Ω g z iθ ~ 9

30 To find θ : Denote g z = Re z z = re iα iθ α = arg g z 1 1 g z g z g z z z Rr e α θ ( ) i( + ) = α + θ =± π θ α π = ± 3

31 y θ α π θ = ± α = arg g z SAP x The direction of integration determines The sign. The user must determine this. 31

32 Summary π I Ω f z e e Ω g z Ω g z iθ ~ α π θ = ± ( ) α = arg g z Ω g z I f z e dz Ω = C 3

33 Example J 1 Ω = π π / π / cos Ωcos z dz 1 = Re I Ω π where π / π / ( icos z) Ω I e dz Ω = Hence, we identify: f ( z) = 1 g z = icos z g z = isin z = z = ± nπ 33

34 Example (cont.) y π z = π x g( z) = icos z g z = icos z = i π α = arg g ( z ) = α π π π θ = ± = ± 4 π 3π θ = or

35 Example (cont.) Identify the and SAP: = cos( + ) = i[ cos xcosh y isin xsinh y] g z i x iy uxy, = sin xsinh y vxy, = cos xcosh y and SAP: v( z) = v( z ) = constant cos xcosh y = constant = cos cosh = 1 35

36 Example (cont.) and SAP: cos xcosh y = 1 y SAP Examination of the u function reveals which of the two paths is the. π C π x uxy, = sin xsinh y 36

37 Example (cont.) Vertical paths are added so that the path now has limits at infinity. y = C + C v1 + C v It is now clear which choice is correct for the departure angle: C v1 C π π C v x θ π = 4 37

38 Example (cont.) π I( Ω) f ( z ) e e Ω g Ω g z iθ ~ ( z ) (If we ignore the contributions of the vertical paths.) Hence, π I( Ω) ~ ( 1) e e Ω i π Ωicos( i ) 4 I ( Ω) ~ π e Ω π i Ω 4 so J π 1 π i Ω Ω π Ω 4 ~ Re e 38

39 Example (cont.) Hence J ( Ω) π ~ cos Ω π Ω 4 39

40 Example (cont.) Examine the path C v1 (the path C v is similar). y C v1 π π x ( i z) π / Ω cos v1 = π / + i I e dz Let z π = + iy 4

41 Example (cont.) π iωcos + iy Ωsinh y Ωsinh y Iv 1 = i e dy = i e dy = i e dy since π cos + iy = sin( iy) = i sinh y Since Ω is becoming very large, we can write: y Iv 1 i e Ω dy 1 Ωy i e Ω Hence, we have 1 Iv 1 i Ω 41

42 Example (cont.) Hence, 1 I = O v1 Ω Hence, I v1 is a much smaller term than what we obtain from the method of steepest descent, when Ω gets large, and we can ignore it. 4

43 Example The Gamma (generalized factorial) function: ( x) ( x 1) t x 1 Γ( x) e t dt, x > Γ =! = Gamma function (generalized factorial function) t Ω t Ω t Ωln t e t dt e e dt e e dt ( ln t) Γ( Ω+ 1) = = = Let t = Ωz Note: There is no saddle point! Ωln Ω Ωz Ωln z Ω Ωz Ωln z e e e dz e e dz Γ( Ω+ 1) = Ω =Ω Ω 43

44 Example (cont.) Ω+ 1 Ωln z Γ( Ω+ 1) =Ω e e dz or Ωz f Ω+ 1 ( z) = Ω ln g z = z z 1 g ( z ) = 1 z Ω+ 1 Γ( Ω+ 1) =Ω e Ω ( ln z z) dz ( z ) z = 1 g z = 1 1 g = = 1 z ( ) α= arg g z = π α π π π θ = ± = ± 44

45 Recall: Example (cont.) π I( Ω) f ( z ) e e Ω g Hence Ω g z iθ ~ ( z ) π Γ( Ω+ 1) Ω e e i Ω or Ω+ 1 Ω Ω Γ( Ω+ 1) π ΩΩ e Ω Note: The is the positive real axis (see the derivation below). The departure angle is zero, not π, since we are integrating from to. ln g z = z z Let Then z = re iθ (, ) vxy : θ y = = θ y This is Sterling s formula. 45

46 Complete Asymptotic Expansion We can obtain the complete asymptotic expansion of the integral with the steepest-descent method (including as many higher-order terms as we wish). Define: b Ωg( z) I Ω = f z e dz a s g z g z h s f z s dz s ( ) ds Then we have Ωg z b Ωs I Ω = e h s e ds sa s 46

47 Complete Asymptotic Expansion (cont.) Ω g z Ωs I e h s e ds Ω (Extending the limits to infinity gives an exponentially small error: This does not affect the asymptotic expansion.) Assume n h s ~ as as s n=,1,,3... n This means : N n N n = h s as o s n= as s Then Ωg z n s n n=,1,,3... Ω I e a s e ds Ω Note: The integral is zero for n = odd (the integrand is then an odd function). This result is called Watson s Lemma : We can obtain the complete asymptotic expansion of I by substituting in the complete asymptotic expansion for h and integrating term-by-term. 47

48 Complete Asymptotic Expansion (cont.) Performing the integration, g( z ) a n + 1 Ω Γ ( n 1) n=,,4 Ω n Ωs s e where We then have 1 n + 1 ds = Γ ( n+ 1) Ω ( x) ( x 1) Γ =! t x 1 Γ( x) e t dt n I ~ e Ω + Use ( n+ ) t s, dt s ds =Ω =Ω 1 Ωs n t n e s ds = e s dt Ω( s) n/ t t 1 e = dt Ω t Ω Ω 1 t ( n 1 )/ = e t dt 1 Ω 1 n + 1 = Γ ( n+ 1) Ω 48

49 Complete Asymptotic Expansion (cont.) Summary b Ωg( z) I Ω = f z e dz a g( z ) a n + 1 Ω ~ Γ ( 1) n=,,4 (even) Ω n I e Ω n+ n h s ~ as as s n=,1,,3... s = g z g z h s = f z s n dz s ( ) ds 49

50 Complete Asymptotic Expansion (cont.) Example: 1 Ωz I z e dz We have cos Ω = 1 g ( z) = z z = = cos f z z = z g z = = ( + ) = ( + ) + ( ) g z z x iy x y i xy v x, y = xy : v x, y = v x, y = xy = : y = (real axis) SAP: x = (imaginary axis) 5

51 Complete Asymptotic Expansion (cont.) Also, we have dz s h( s) = f ( z( s) ) ds s = g z g z = z + z z= = + z = Hence s = ± z ( must choose + sign) z s < s > 1 1 We then have dz ds f ( z( s) ) = cos( z( s) ) = cos( s) = = 1 ds ds y z = x 51

52 Complete Asymptotic Expansion (cont.) Recall: Hence ( s) ( s) dz s h( s) = f ( z( s) ) ds = cos 1 = cos n h s ~ as as s n=,1,,3... n 4 s s = cos 1 + a = 1, a 1 1 =, a4 =, etc.! 4! 4 ( s) h s 5

53 Complete Asymptotic Expansion (cont.) The complete asymptotic expansion is g( z ) a n + 1 Ω ~ Γ ( n 1) n=,,4 Ω Hence we have: n I e Ω + a 1 a 3 Ω Γ + Γ + 3/ Ω Ω I ~ e Ω so that I 1 1 1/ 3 Ω Γ + Γ + Ω Ω ~ 3/ 1 1 a = 1, a =, a4 =, etc. 4 where 1 Γ = π 3 1 Γ = π (Please see the notes on the Gamma function.) 53

54 Complete Asymptotic Expansion (cont.) Hence we have: 1 Ωz π 1 π 3/ Ω 4 Ω 1 cos I Ω = z e dz + as Ω Note: In a similar manner, we could obtain higher-order terms in the asymptotic expansion of the Bessel function or the Gamma function. 54