Notes 7 Analytic Continuation

Transcription

1 ECE 6382 Fall 27 David R. Jackson Notes 7 Analtic Continuation Notes are from D. R. Wilton, Dept. of ECE

2 Analtic Continuation of Functions We define analtic continuation as the process of continuing a function off of the real ais and into the comple plane such that the resulting function is analtic. More generall, analtic continuation etends the representation of a function in one region of the comple plane into another region, where the original representation ma not have been valid. For eample, consider the Bessel function J n (). How do we define J n () so that it is computable in some region and agrees with J n () when is real? 2

3 Analtic Continuation of Functions (cont.) One approach to etend the domain of a function is to use Talor series. We start with a Talor series that is valid in some region. We etend this to a Talor series that is valid in another region. Note: This ma not be the easiest wa in practice, but it works. 3

4 Analtic Continuation of Functions (cont.) two alternative representations f ( ) n = =, < n= Epand about =. Since both series are ( ) = m ( + 2 ) f b 2 valid there, find coefficients of new series b differentiating the original series : where m= m, + < use the above series representation Eample b n n n n m m m d f d n n m m = = ( )( 2) ( m m = + ) m! d m! d n m! = = = n= m = 2 2 2! ( ) ( ) m n n m d n = 2 ( Note =, n =,,,m ) m m! n= m n m! d The coefficients of the new series --- with etended region of convergence --- are determined from the coefficients of the original series, even though that series did not converge in the etended region. The information to etend the convergence region is contained in the coefficients of the original series --- even if it was divergent in the new region! 2 4

5 Analtic Continuation of Functions (cont.) Eample (cont.) Another wa to get the Talor series epansion: m m 3 d m, m m m= m! d = 2 / ( ) ( ) f = b + + < b = so that b m 2 = = m! 3 ( )( 2)( 3) ( m) ( ) m= m+ m+ = 2 / 2 f ( ) = ( + ) + < 3 m, m+ Note: This is sort of cheating in the sense that we assume we alread know a closed form epression for the function. 5

6 Analtic Continuation of Functions (cont.) Eample (cont.) f ( ) = Here we show the continuation of f ( ) from its power series representation in the region < into the entire comple plane using Talor series. 2 If the singularities are isolated, we can continue an function into the entire comple plane via a sequence of continuations using Talor and / or Laurent series! 6

7 The Zeros of an Analtic Function are Isolated Proof of theorem : (The eros cannot be arbitraril close together) Assume that f( ) is analtic in a connected region A, and suppose f( ) =. Then f( ) has a Talor series f( ) = a ( ) + a ( ) + with a = f( ) = at. 2 2 More generall, we ma have a ero with multiplit n such that we also have f ( ) = = f ( ) =, f ( ). In this case, f( ) has a Talor series ( n ) ( n) n n+ = n + n+ + f( ) a ( ) a ( ) n ( n) = ( ) an + an+ ( ) + ( an = f ( ) ) n! n = ( ) g ( ), g ( ), and g ( ) is analtic in A(it is represented b a converging Talor series). Since g ( ) is analtic it is also continuous. Since g ( ), g ( ) cannot vanish within a sufficientl small neighborhood of. That is, the eros of f( ) must be isolated. The onl eception is if the function f () is identicall ero. 7

8 The Zeros of an Analtic Function are Isolated Eample : The function f( ) = sin ( ) has eros at =, n, 2, nπ = but it cannot be analtic at = since the eros accumulate there and hence are not isolated. / π The origin is an accumulation point for the eros. 8

9 Analtic Continuation Principle Theorem of analtic continuation: Assume that g() is analtic in a connected region A, and it equals an analtic function f () on a set of distinct points n in A that converge to a point in A. (That is, f ( n ) = g( n ).) Then g() = f () in A. In other words, there can be onl one function that is analtic in A and has a defined set of values at the converging points n. n A g() Note: Defining n on a contour in A also suffices. Proof: Construct the difference function g()-f(), which in analtic in A. This function must have a Talor series at. This function has all ero coefficients; otherwise, the analtic difference function must have isolated eros which it does not, b assumption. B continuing the Talor series that has ero coefficients (analtic continuation), the difference function must be ero throughout A. 9

10 Analtic Continuation Principle (cont.) Corollar (etending a domain from A to A B) Assume that f () is analtic in A and g() is analtic in B, and the two domains overlap in a region A B, and f = g in A B. Define h( ) ( ) ( ) f, A = g, B Then h() is the onl analtic function in A B that equals f () on A. f ( ) = g( ) A A B A B B Proof: The function h is analtic in the region A B and equals f () on an set of converging points in the intersection region. The theorem of analtic continuation thus ensures that h is unique in A B.

11 Analtic Continuation Principle (cont.) Eample (eample of theorem) The function sin () is continued off the real ais. ( ) sin ( ) g = e j e 2 j j Line segment ( ) = sin ( ) f The function g() is the onl one that is analtic in the blue region of the comple plane and agrees with sin () on an part of the real ais.

12 Analtic Continuation Principle (cont.) Eample (eample of theorem) The Bessel function J n () is continued off the real ais. ( ) g k= ( ) ( k) k n+ 2k = k! n+! 2 Line segment k= ( ) ( k) k n+ 2k f ( ) = Jn ( ) = k! n+! 2 The function g() is the onl one that is analtic in the blue region of the comple plane and agrees with J n () on an part of the real ais. 2

13 Analtic Continuation Principle (cont.) Eample (of corollar) The function Ln () (principal branch) is continued beond a branch cut. ( ) Ln ( ) ln ( ) f = r + i θ, π < θ < π Note: Ln denotes the principal branch. Region of overlap (second quadrant) Branch cut for f Branch cut for g ( ) ln ( ) ln ( ) g = r + iθ π / 2< θ < 3π / 2 In the blue region (second and third quadrants), g() is analtic and agrees with the function f () in the second quadrant. The original function f () is not analtic in the entire blue region. 3

14 Analtic Continuation Principle (cont.) Eample (cont.) The function g() is the onl function that is analtic in the entire left-half plane and agrees with Ln() in the second quadrant. ( ) ( ) ( ) g ln = ln r + i θ, π / 2 < θ < 3π / 2 4

15 Analtic Continuation Principle (cont.) Eample (of corollar) The function Y n () (Bessel function of the second kind) is continued beond a branch cut. Branch cut ( n k ) n 2k n! γ π k= k 2 Yn( ) = Jn( ) Ln π + 2! 2 k ( ) ( k) ( n k) π Φ +Φ + k= k! n+ k! 2 ( ) Φ ( p) = ( p> ) 2 3 p Ln ( ) = ln ( ) + π < θ < π r iθ 2k+ n 5

16 Analtic Continuation Principle (cont.) Eample (cont.) The etended function is analtic within the blue region. Note: In the third quadrant, this etended function will not be the same as the usual Bessel function there. In the third quadrant : et Y ( ) = Y ( ) n n 2 + Jn i π ( )( 2π ) ln ( ) ln ( ) = r + iθ π / 2< θ < 3π / 2 ( n k ) n 2k n! γ π k= k et 2 Yn ( ) Jn( ) ln π + 2! 2 k ( ) ( k) ( n k) π Φ +Φ + k= k! n+ k! 2 ( ) 2k+ n 6

17 Analtic Continuation Principle (cont.) Eample How do we etend F() to arbitrar? t ( ) ( ) F e J t dt, Original domain: > Identit : t e J ( t) dt =, 2 + 7

18 Analtic Continuation Principle (cont.) We define: Eample (cont.) ( ) F ( 2 + ) 2 / This is defined everwhere in the comple plane (ecept on the branch cuts). Note: The shape of the branch cuts is arbitrar. i i ( ) 2 / = + 8

19 Analtic Continuation Principle (cont.) Eample How do we etend F() to arbitrar? Im ( t) ( ) F dt t Original domain: > Re( t) ( ) = ln ( ) ln ( ) = ln ( ) arg ( ) : Choose θ = ( ) Ln ( ) F ln = = real number Recall: ( ) means π ( ) Ln < arg < π 9

20 Analtic Continuation Principle (cont.) Eample (cont.) Im ( t) ( ) F dt t C Re( t) We analticall continue F() from the real ais. We require that the path is varied continuousl as leaves the real ais. 2

21 Analtic Continuation Principle (cont.) Eample (cont.) Im ( t) C + ( ) F dt t F + ( ) F ( ) C Re( t) As we encircle the pole at the origin, we get a different result for the function F(). + d = d d = F ( ) F ( ) = 2π i + C C C 2

22 Analtic Continuation Principle (cont.) Branch cut plane Eample (cont.) F dt t ( ) = Ln ( ) C The function F() is single-valued and analtic as long as we do not encircle the origin. The branch cut keeps us from doing this. 22

23 Analtic Continuation Principle (cont.) Eample How do we interpret F() for arbitrar? ( ) ( ) ( ) 2 cosh = Ln + F Note : > Note : ( ) ( ) 2 2 is real ( ) Ln + arg + = Original domain: > Recall: ( ) means π ( ) Ln < arg < π 23

24 Analtic Continuation Principle (cont.) Eample (cont.) cosh ln / F( ) = ( ) = + ( 2 ) 2 Two issues: () The square root function (2) The ln function ( ) ( ) ( ) 2 cosh = Ln + F Both functions must be varied continuousl from the real ais. 24

25 Analtic Continuation Principle (cont.) Choose : ( ) ( ) 2 / 2 2 f = = ln = Ln Eample (cont.) ( ) ( ) ( ) 2 = cosh = Ln + F Note: The function f () is discontinuous as we cross the branch cuts. Sommerfeld branch cuts F() is analtic in an one quadrant. 2 ( ) ( ) f = Re f > Sommefeld branch cuts ( Recall : Re > ) 25

26 Analtic Continuation Principle (cont.) Do we need another branch cut due to the Ln function? Eample (cont.) ( ) ( ) ( ) 2 = cosh = Ln + F Eamine argument of Ln function: 2 w = + = w+ 2 w ( derivation omitted) The branch cut for the Ln function corresponds to w being a negative real number. ( ) ( ), w, 26

27 Analtic Continuation Principle (cont.) Eample (cont.) ( ) ( ) ( ) 2 = cosh = Ln + F F() is an analtic continuation of cosh - () off of the real ais. F() is analtic in an one quadrant. 27

28 Analtic Continuation Principle (cont.) Eample (cont.) Another choice of branch cuts F () is now analtic in the upper half plane. The branch cut from = - is chosen to lie outside the second quadrant. θ 2 2 ( ) = ( ) ( + ) 2 / 2 / 2 / The angles θ and θ 2 change continuousl. ( ) θ 2 2 / 2 = ( θ = θ = ) 2 2 F( ) = cosh ( ) = Ln + ( ) 2 / Note: The radical sign is not used an more! 28

29 Analtic Continuation Principle (cont.) Eample (cont.) Quadrant #2 : ( ) 2 ( ) 2 / Re < 2 2 / 2 Quadrant #: ( ) 2 / 2 2 ( ) 2 / 2 2 = 2 F( ) = cosh ( ) = Ln + ( ) 2 / F () is now analtic in the upper half plane. 29

30 EM Eample Assume a radiating phased line source on the ais. Note: j is used here instead of i. k is real I( ) = I e jk The magnetic vector potential is µ I = 4 j (2) A H ( kρ ρ) e jk 2 ( 2) /2 k k k ρ = 2 2 k k, k < k = j k k, k > k 2 2 3

31 EM Eample (cont.) We can also write (,, ) jkr jk e A = I e µ d 4 π R Note: The integral converges for real k. R I( ) = I e jk (,, ) Hence jkr jk e µ I (2) A = Ie µ d = H ( kρ ρ) e 4π R 4j jk Eists onl for real k Eists for comple k The second form is the analtic continuation of the first form off of the real ais. 3

32 EM Eample (cont.) µ I = 4 j (2) A H ( kρρ) e 2 ( 2 /2 ) 2 2 j( k k ) k k k ρ = = /2 jk k is comple I( ) = I e jk In order for this to be the analtic continuation off of the real ais of the integral form, we must chose the branch of the square root function correctl so that it changes smoothl and it is correct when k = real > k. 2 2 ( ) /2 k = j k ρ k = negative imaginar number for k = real > k 32

33 EM Eample (cont.) ( 2 2) /2 k = ρ j k k Im ( k ) k = ρ j k k 2 2 Im k ρ < Im k ρ < k Re ( k ) k Im k ρ < Im k ρ < The Sommerfeld branch cuts are a convenient choice. 33

34 EM Eample (cont.) ( 2 2) /2 k = ρ j k k 2 2 Im ( k ) k = ρ j k k (Top sheet) Riemann surface k Re ( k ) k Note: The function is continuous on the Riemann surface. We can now let k wander anwhere we wish on the Riemann surface, and we know how to calculate the square root. 34

35 EM Eample (cont.) Eample What is k ρ at the final indicated point? ( 2 2) /2 k = ρ j k k Im ( k ) k = ρ j k k 2 2 (Top sheet) Riemann surface k Im k ρ < Im k ρ > k (Bottom sheet) Re ( k ) At the indicated final point, the imaginar part of k ρ is chosen to be positive. 35

36 EM Eample (cont.) Eample What tpe of wave is at the indicated points? ( 2 2) /2 k = ρ j k k Im ( k ) Surface wave : ( ) k = ρ j + Riemann surface Phased arra : k ρ = ( + ) k Re ( k ) k Leak wave : ( ) j( ) k = ρ Note : ( ) /2 /2 ( ) ( ) k j k k k k = ρ Note: The leak wave field is the analtic continuation of the phased arra field when k becomes comple. 36

37 Schwar Reflection Principle Assume that f () is the analtic continuation of a real function f () off the real ais (or a segment of the real ais). f = f Then within the analtic region, we have ( * ) ( ) * (proof omitted) Line segment: f f ( ) ( ) Eamples: ( ) e J ( ) sin,, ln ( ), Re( ) > n f () is assumed analtic in this region. (assuming branch cut on negative real ais) 37