prof. david r. jackson ece dept. spring 2014 notes 33 ece 6341 1

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Prof. David R. JacksonECE Dept.

Spring 2014

Notes 33

ECE 6341

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Steepest-Descent MethodComplex Integral:

Peter Joseph William Debye (March 24, 1884 – November 2, 1966) was a Dutch physicist and physical chemist, and Nobel laureate in Chemistry.

http://en.wikipedia.org/wiki/Peter_Debye

The method was published by Peter Debye in 1909. Debye noted in his work that the method was developed in a unpublished note by Bernhard Riemann (1863).

This is an extension of Laplace’s method to treat integrals in the complex plane.

Georg Friedrich Bernhard Riemann (September 17, 1826 – July 20, 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.

http://en.wikipedia.org/wiki/Bernhard_Riemann

g z

C

I f z e dz

http://en.wikipedia.org/wiki/File:Georg_Friedrich_Bernhard_Riemann.jpeg

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Steepest-Descent Method (cont.)

Complex Integral:

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

C

I f z e dz

0 0g z

0g

x

0

g

y

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).

pC C ' C

2 Resp

p

C

F z dz j F z

C

C

x

y

0z

X

pCpz

5


Denote

Cauchy Reimann eqs.:

g z u z j v z

u v

x y

u v

y x

Hence2

2

2

2

u v v

x x y y x

u

y

6


or

2 2

2 20

u u

x y

0,xxu

2

0 0 0

2

0 0 0

1, ,

21

2

xx

yy xy

u x y u x y u x x

u y y u x x y y

Ignore (rotate coordinates to eliminate)

Near the saddle point:

0yyu If then

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2 2

0 0 0 0 0 0

1 1, ,

2 2xx yy xyu x y u x y u x x u y y u x x y y

2 2

0 0 0 0

1 1, ,

2 2x x y yu x y u x y u x x u y y

In the rotated coordinate system:

Assume that the coordinate system is rotated so that

0x xu 0y yu

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Steepest-Descent Method (cont.)The u (x, y ) function has a “saddle” shape near the SP:

x

y

( , )u x y

0z

9


Note: The saddle does not necessarily open along one of the principal axes (only when uxy (x0, y0) = 0).

y

( , )u x y

0zx yx

10


Along any descending path (where the u function decreases):

00

0 00

0

0

~g z g zg z

C

j v z v z u z u zg z

C

I f z e e dz

f z e e e dz

Both the phase and amplitude change along an arbitrary path C.

If we can find a path along which the phase does not change (v is constant), the integral will look like that in Laplace’s method.

11


Choose path of constant phase:

0 0:C v z v z constant

y

0z

x

0C

12


Hence C0 is either a “path of steepest descent” (SDP) or a “path of steepest ascent” (SAP).

0,u x y C is parallel to

Gradient Property (proof on next slide):

SDP: 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.

13


Hence,

0

u v u vu v

x x y y

u u u u

x y y x

u v

Hence0 u C

Proof

ˆ ˆ

ˆ ˆ

u uu x y

x y

v vv x y

x y

00 ( )Cv C v is constant onAlso,

v

ux

y

C0

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0 00

0~j v z v z u z u zg z

C

I f z e e e dz

00

0~u z u zg z

SDP

I f z e e dz

Because the v function is constant along the SDP, we have

00

0~g z g zg z

SDP

I f z e e dz

or

real

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Local behavior near SP

20 0 0 0 0

1

2g z g z g z z z g z z z

20 0 0

1

2g z g z g z z z so

0

0

j

j

g z R e

z z re

Denote

220

1

2jg z g z R r e

x

y

r

z0

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220

1

2jg z g z Rr e

SDP:

2 0 2 n SAP:

2 2 n

,2 2

,2 2 2 2

0( ) ( ) 0v z v z

0( ) ( ) 0v z v z

20

1( ) ( )

2u z u z Rr

20

1( ) ( )

2u z u z Rr

Note: The two paths are 90o apart at the saddle point.

20

1cos 2

2u z u z Rr

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SAP

SDPSP

x

y

u decreases

u increases

90o

u

u

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SAP

SDP

y

( , )u x y

0z

x

19


00

0~u z u zg z

SDP

I f z e e dz

Set 20s u z u z This defines

( )z z s

y

0s x 0s

,u x y

SDP0z

2s

z

20


Hence

20

0~ g z s dzI f z e e ds

ds

20

00

~ g z s

s

dzI f z e e ds

ds

0

00

~ g z

s

dzI f z e

ds

(leading term of the asymptotic expansion)

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Take one more derivative:

To evaluate the derivative:

20

0

s u z u z

g z g z

2

22 2

ds ds d ss g z

dz dz dz

At the SP this gives 0 = 0. 2ds

s g zdz

(Recall: v is constant along SDP.)

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At 0 :s 0

1

20

2Z

g zds

dz

so

1

2

0 0

2

s

dz

ds g z

0

1

2

00

2~ g zI f z e

g z

Hence, we have

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Note: There is an ambiguity in sign for the square root:

0

arg SDPs

dz

ds

1

2

0 0

2

s

dz

ds g z

To avoid this ambiguity, define

y

SDP

x

0z

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Hence

0 0

2SDPj

s

dze

ds g z

The derivative term is therefore

0

00

2~ SDPg z jI f z e e

g z

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To find SDP :

0

0SDP

j

j

g z R e

z z r e

Denote

2

2 2

SDP

SDP

2 220 0 0

1 1

2 2SDPjg z g z g z z z Rr e

0arg g z

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2 2SDP

Note: The direction ofintegration determinesThe sign.

The “user” must determine this.

SDP

SAP

y

SDP

x

0arg g z

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Steepest-Descent Method (cont.)Summary

2 2SDP

0arg g z

0

00


g z

g z

C

I f z e dz

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Example

/ 2

0 / 2

1cos cos

1Re

J z dz

I

1

cos

f z

g z j z

0 0sin 0g z j z

/ 2 cos

/ 2

j zI e dz

where

Hence, we identify:

0 0z n

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Example (cont.)

0 0

0

cos

arg2

2 2 4 2SDP

g z j z j

g z

cosg z j z

3

4 4SDP

or

2

2

x

y

0 0z

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Example (cont.)

SDP and SAP:

cos

cos cosh sin sinh

, sin sinh

, cos cosh

g z j x jy

j x y j x y

u x y x y

v x y x y

0v z v z constant

cos cosh cos 0 cosh 0

1

x y

constant

Identify the SDP and SAP:

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Example (cont.)

cos cosh 1x y

, sin sinhu x y x y

Examination of the u function reveals which of the two paths is the SDP.

SDP and SAP:

2

SAP

SDP

y

Cx

2

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Example (cont.)

Vertical paths are added so that the path now has limits at infinity.

SDP = C + Cv1 + Cv2

4SDP

It is now clear which choice is correct for the departure angle:

SDP

1vC

x

2

2

2vC

C

y

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Example (cont.)

(If we ignore the contributions of the vertical paths.)

cos 0 4

4

2~ 1

2~

jj

j

I e ej

I e

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1 2~ Re

j

J e

so

0

00


g z

Hence,

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Example (cont.)

Hence

0

2~ cos

4J

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Example (cont.)

/2cos

/21j z

jv e dzI

2z jy

Examine the path Cv1 (the path Cv2 is similar).

Let

y

1vC

x

2

2

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Example (cont.)

0 0cos

sinh sinh21

0

j jyy y

vI j e dy j e dy j e dy

sinh1

0

sinh

0

1

cosh

1

cosh

yv

y

dI j e dy

y dy

j ey

1

1v

I j

Use integration by parts (we can also use Watson’s Lemma):

cos sin( ) sinh2

jy jy j y

since

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Example (cont.)

1

1v

I O

Hence,

If we want an asymptotic expansion that is accurate to order 1/ , then the vertical paths must be considered.

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Example (cont.)

Alternative evaluation of Iv1 using Watson’s Lemma (alternative form):

sinh1

0

yv

I j e dy

sinh

cosh

s y

dsy

dy

Use

10

0

20

20

1

cosh

1

1 sinh

1

1

sv

s

s

s

dyI j e ds

ds

j e dsy

j e dsy

j e dss

2

10

12

sv

sI j e ds

Applying Watson’s Lemma:

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Example (cont.)2

10

12

sv

sI j e ds

21

0 02

s sv

jI j e ds e s ds

1 3

312v

jI j

Hence,

so

10

11n

n

sns e ds

Recall:

1

01 ! x tx x t e dt

Note: Iv1 is negligible

compared to the saddle-point contribution.

(Watson’s lemma)

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Complete Asymptotic Expansion

By using Watson's lemma, we can obtain the complete asymptotic expansion of the integral in the steepest-descent method, exactly as we did in Laplace's method.

0

10,2,4 2

1~

2g z n

nn

a nI e

b g z

aI f z e dz

dz sh s f z s

ds

0,2,4...

~ 0nn

n

h s a s s

as

20 SDP

s g z g z

prof. david r. jackson ece dept. spring 2014 notes 33 ece 6341 1

Documents

path of steepest descent

debyethe method

extension of laplaces

path of steepest ascent

path deformation

descending path

path of constant phase

saddle point sp