Global Theory of a Second Order Linear Ordinary Differential Equation With a Polynomial Coefficient (North-Holland mathematics studies 18)

,

,

9 p= Z n j =1 j '

where

Furthermore, he showed t h a t such a covering surface R equivalent t o the complex plane. R

x-plane onto function

f

.

z = f(x)

Let

i s conformally

be a conformal mapping from

Then, Nevanlinna a l s o showed t h a t t h e Schwarzian of the

i s a polynomial i n

x

of degree

zation of t h e d i f f e r e n t i a l equation

(A)

p-2

.

This is a characteri-

i n terms of covering surfaces of

I n Chapter 8, an i n t e r p r e t a t i o n of Nevanlinnafs work

the Riemann sphere.

i s given by the following r e s u l t :

Let co,. ( a f f ) ck # ck+l (b")

the s e t

..,c mtl

be given points on t h e Riemann sphere.

..,mtl; cmt2 =

(k= 0,. [co,.

..,c

IUtl

CJ

7

Assume t h a t

and

contains a t l e a s t t h r e e d i s t i n c t p o i n t s on

)

t h e Riemann sphere.

5 ,...,am

Then, t h e r e e x i s t

such t h a t the d i f f e r e n t i a l equation

two l i n e a r l y independent s o l u t i o n s y,(x)

y,(x)

(A)

has

s a t i s f y i n g the con-

ditions

Y+) y2(x)

-+

"k as

The converse i s a l s o t r u e .

x-)-

in %

.

(k = 0 , . .,mtl)

.

( C f . Theorem 39.1, Section 39.)

Allen Weitsman and David Drasin of Purdue University showed t h e author the importance o f Nevanlinna 1s contribution. I n Chapter 9, a converse problem i s solved with regard t o t h e connect i o n formulas

(D)

by u t i l i z i n g the r e s u l t s of Chapters 5 and 8.

A more

general problem which was proposed by G. D. Birkhoff 1.4 and 5 1 i s a l s o summarized. Chapters 2 and 7 a r e based on the a u t h o r ' s j o i n t works with Po-Fang Hsieh of Western H c h i g a n University. Mullin of TRW, Washington D . C . .

Chapter 3 explains a work of F'rank

The content of Section 26 of Chapter 5 i s

based on a work o f Herman Gollwitzer of Drexel University.

48 of Chapter 8 summarize a recent work of I v a r -en

Sections 47 and

of Norway.

Some

INTRODUCTION

X

p o r t i o n s of Chapters 5 , 9 and 10 have never been published b e f o r e .

in 1958, t h e a u t h o r worked o u t a s i m p l i f i c a t i o n o f a second o r d e r linecir orciinary c i i i f e r e n t i a l e q u a t i o n a t a t r a n s i t i o n p o i n t .

'The m o t i v a t i o n was t o g e n e r a l i z e t h e r e s u l t s of R.E.

Sibuya [ 3 2 ) , )

Y. Langer

YcKelvey [ 2 3 ] by u t i l i z i n g a method based on t h e g e n e r a l i d e a

[ 2 2 ! and R.W.

of' b!.

(Cf.

Wcuhara.

An e x c e l l e n t guidance f o r t h i s r e s e a r c h was a l s o found i n

m e x p o s i t o r y paper by W . Wasow [ 4 8 ] .

As t h e main r e s u l t , a t r a n s f o r m a t i o n

w a s constructed s o t h a t t h e g e n e r a l case was reduced t o t h e d i f f e r e n t i a l equation parameter. n = l rmd

(A)

, where

the coefficients

5 , ...,a m

a r e f u n c t i o n s of another

'fie r e s u l t s o? Langer and McKelvey were reproduced as t h e c a s e s m=2

respectively.

mation was n o t c l e a r f o r

However, t h e u s e f u l n e s s of such a t r a n s f o r -

a23 owing t o t h e f a c t t h a t t h e i n f o r m a t i o n of

t h e g l o b a l behavior o f s o l u t i o n s o f t h e d i f f e r e n t i a l e q u a t i o n easily available for

m23 .

(A) was n o t

Thus, t h e a u t h o r was motivated toward t h e

r e s e a r c h concerning t h e d i f f e r e n t i a l equation

(A)

.

A s u b s t a n t i a l improve-

ment of the r e s u l t mentioned above was obtained i n 1973 as an a p p l i c a t i o n o f t h e r e s u l t s which had been accumulated through t h i s r e s e a r c h .

Si'mya [ 4 0 ] .)

( C f . Y.

These s p e c i a l t o p i c s are n o t included i n t h e p r e s e n t book.

The r e s e a r c h of t h e a u t h o r has been guided and encouraged by P r o f e s s o r b s u o ' M a r a and many o t h e r g r e a t t e a c h e r s and f r i e n d s .

For t h e b a s i c

guide-line and many u s e f u l i n f o r m a t i o n s concerning this r e s e a r c h , t h e a u t h o r i s indebted t o P r o f e s s o r Walfgang Wasow.

The a c t i v i t y with Profes-

s o r ' d i l l i a m A . H a r r i s , Jr. has k e p t t h i s r e s e a r c h going i n a v e r y p l e a s a n t

atmosphere

Ir, t h e s p r i n g o f 1969, t h e a u t h o r gave a series of l e c t u r e s on t h e d i f f e r e n t i a l equation

(A)

, at

t h e U n i v e r s i t y of C a l i f o r n i a , Los Angeles.

Then, he s t a r t e d p u t t i n g materials t o g e t h e r i n t o a "bookf1. A p a r t of t h e manuscripi, was w r i t t e n a t t h e U n i v e r s i t y of Edinburgh, S c o t l a n d , i n t h e s p r i n g of 1971.

A complete manuscript was f i n i s h e d a t Mathematics Research

Center, t h e U n i v e r s i t y of Wisconsin-Madison, i n t h e academic year 1972/73, during t h e a u t h o r ' s s a b b a t i c a l l e a v e from t h e U n i v e r s i t y o f Minnesota.

The

manuscript was examined a t l e a s t by P r o f e s s o r Tosihusa Kimura, P r o f e s s o r Masahiro Iwano, Dr. l v a r Bakken, and P r o f e s s o r Walfgang Wasow.

Following

t h e i r advice and comments, t h e manuscript was r e v i s e d s o t h a t t h e book might become more r e a d a b l e .

changed $hree times. author

tG

For example, t h e t i t l e o f this book has been

P r o f e s s o r R E . O'Malley, Jr. k i n d l y i n t r o d u c e d t h e

Dr. E. F'redriksson of North-Holland P u b l i s h i n g Company.

The

xi

INTRODUCTION

camera copy was b e a u t i f u l l y typed by M s . Kathryn L. Nelson.

The author

wishes t o express his s i n c e r e appreciation t o his teachers and f r i e n d s f o r t h e i r guidance, encouragement, advice, comments, and helps.

This research has been supported by the University o f Minnesota, t h e University o f Wisconsin, t h e University of C a l i f o r n i a , t h e National Scienoe Foundation, the United S t a t e s Navy, t h e United S t a t e s Army, and t h e Science Research Council a t the University o f Edinburgh through various f a c i l i t i e s and funds.

YASUTAKA SIBUYA Roseville June 1975

This Page Intentionally Left Blank

CONTENTS Introduction

V

Chapter 1 Asymptotic expansions

.

1 2

..................................

Definition of asymptotic expansions

.

Uniqueness of asymptotic expansions

. 4. 3

Properties o f asymptotic expansions Summation o f formal power series

5.

1

..........................

.......................... .......................... .............................

Asymptotic expansions o f functions depending on parameters

1 2

3 10

...

.................... 6 . An existence theorem ......................................... 7 . Subdominant s o l u t i o n s ......................................... 8 . Case m = l and case m = 2 ................................... 9 . Proof of Theorem 6.1 ......................................... 1 0 . h o o f of Lemma 9.1 ........................................... Chapter 3 A r e l a t i o n between l,jm and l,jmtl ...................... 11. Remarks on perturbations of asymptotic s o l u t i o n s ............. 12 . Main problem ................................................. 13 . Asymptotic s o l u t i o n s a t x = + ............................... 2 1.4. Case when F(x. = e x + x .................................... 15 . A generalization of Zernike's method ......................... 16 . A subdominant s o l u t i o n of equation (15.2) .................... 17 . Proof of Theorem 12.1 ........................................ Chapter 2

D e h t i o n of subdominant s o l u t i o n s

6)

Chapter 4

. 19 . 20 .

Asymptotic behavior o f

um(x.a )

as

a

m

tends t o i n f i n i t y

................................................. ...................... Estimates i n the s e c t o r \arg X 1 I n - po ...................... Estimates i n the s e c t o r larg A - nI 5 po ChaDter 5 Stokes m u l t i p l i e r s ..................................... 21 . Preliminary r e s u l t s .......................................... ................................... 22 . Case m = l and case m = 2 23 . Properties a t a = O ..........................................

18

24

.

25

.

Main p r d l e m

Asymptotic behavior of Stokes m u l t i p l i e r s as infinity Case when

15 15 17 22 26 32 42 42

45 48

51 56 59 61

65 65

66 76 82 82 87

90

am tends t o

..................................................... + a .....................................

P(x, A ) = x3

12

100 101

CONTENTS

XiV

................................................. 27 . A functional equation ........................................ .......... Chapter 6 A boundary value problem i n the complex plane 28 . tioundary conditions .......................................... 29 . D i s t r i b u t i o n of l a r g e eigenvalues ............................ 26

30

.

General case

.

A imique s o l u t i o n o f a non-homogeneous boundary value problem

.

Estimate of

..

................................. 31.. Estimate of R ( b ) g : P a r t I1 ................................ 33 . P r i n c i p a l p a r t s of R(X)g a t i t s poles ...................... 34 . A completeness property of eigenfunctions .................... Chapter 7 D i s t r i b u t i o n of zeros .................................. 35 . A r e l a t i o n between zeros and eigenvalues ..................... 30 . Main theorem ................................................. 37 . Proof of Theorem 36.1: P a r t 1 ............................... 38 . Proof o f ‘Beorem 36.1: P a r t 11 .............................. 31

3iapter 8

R(A)g : P a r t I

.

.

40

41 .

. 43 .

42 &$

.

. 46 . 47 . 48 .

45

..................................................... A general boundary v a l u e problem ............................. Associated fliemann s u r f a c e s .................................. Construction o f a covering s u r f a c e P ........................ Remarks on non-uniqueness and s i n g u l a r i t i e s of s o l u t i o n s ..... A remark on t h e g e n e r a l c a s e ................................. An a p p l i c a t i o n of a l i n e complex ............................. Proof of Theorem 40.1 ........................................ Proof of Theorem 39.1 ........................................ Local uniqueness ............................................. A sketch of t h e proof of Theorem 47.1 ........................

Chapter

2

50 .

51

.

. 53 . 52

54

55

. .

128 128 129

133 140

14.4 147

149 152 152

154 156

161

166 166 172

176 181 185

186

191 1% 197

198

Subdominant s o l u t i o n s admitting a prescribed Stokes

................................................... Main problems .................................................. Case m = l and case m = 2 ................................... Problem 49.1 a t a = O ........................................ Problem 49.2 a t a = O ........................................ Problem 49.1: t h e general case .............................. Problem 49.2: t h e general case .............................. phenomenon

49 .

122

A general boundary value problem and a s s o c i a t e d Riemann

surfaces

39

112

Subdominant s o l u t i o n s admitting a prescribed Stokes phenomenon

201 201 203 204 207 211 215 220

xv

CONTENTS

56

.

A g e n e r a l problem of G.D.

Birkhoff

...........................

225

....

Chapter 10 Subdominant s o l u t i o n s of t h e d i f f e r e n t i a l equation 2 y" .a ( x - x ~ ) (x.x, ) (x - x ~ y ) =0

... 57 . Subdominant s o l u t i o n s ........................................ 58 . Formal s o l u t i o n s of t h e a s s o c i a t e d R i c c a t i equation .......... 59 . Asymptotic s o l u t i o n s i n a canonical domain as X tends t o infinity ..................................................... 60

.

. 62 .

61

Asymptotic s o l u t i o n s i n a canonical domain as infinity Fkamples of

x

..................................................... Stokes curves and canonical domains .............. h

.................................................. 63 . Simple t r a n s i t i o n p o i n t problems i n unbounded domains ........ .................................. 6 4 . A method due t o T.M. Cherry 65 . Asymptotic s o l u t i o n s i n t h e domain Bo ....................... 66 . Remarks on t h e case when n c o n t a i n s o n l y one o f t h e t h r e e 5

and

xo

242

249 253 261 266 269 272

62 i n t h e neighborhood of t h e simple

.......................................... 67 . Examples ..................................................... ........................................................ References Index ............................................................. transition point

236

tends

to infinity

0'

233

tends t o

Asymptotic r e p r e s e n t a t i o n s of subdominant s o l u t i o n s as

sectors 6

233

279 281 285 289

This Page Intentionally Left Blank

CHAPTER 1 ASYMPTOTIC EXPANSIONS

1. Definition of as.ymptotic expansions. B

L e t us consider a closed s e c t o r

i n x-plane which i s defined by

where

,

8: a l a r g x l b

(1.1)

a

and

b

d e r a function

1x1 2

M ,

are realnumbers", and

f(x)

M i s a p o s i t i v e number.

which i s holomorphic i n

be a formal power series i n

x

-1

B

.

Consi-

Let

whose c o e f f i c i e n t s a r e complex numbers.

D e f i n i t i o n 1.1: We c a l l t h e formal series (1.2) an asymptotic expansion

of

f(x)

N

,

as

x

tends t o i n f i n i t y i n

B if, f o r every non-nepative i n t e -

t h e r e exists a p o s i t i v e number

%

such t h a t

(1.3)

for

xEB

.

A closed s e c t o r

open s e c t o r 8 '

(defined by (1.1))i s c a l l e d a subsector of an

defined by aI<arg x
(1.4)

,

1x1 > M I

,

if

at
(1.5)

,

M>M1

.

1.2: We c a l l t h e formal s e r i e s (1.2) an asymptotic expansion of f(x)

x

tends t o i n f i n i t y i n an open s e c t o r

fil

(defined by (1.4))if

the formal series (1.2) i s en asymptotic emansion of

.

f(x)

x

tends

* b - a may be l a r g e r than 2n If b - a > 2n , the domain &Imust be regarded a s a domain on the Riemann surface determined by l o g x

.

ASYMPTOTIC EXPANSIONS

2

t o i n f i n i t y i n every closed subsector of

gl

.

W e s h a l l use the notation

(1.6)

z

f(x)

cNx-N

(x+m

in

Q)

N=o t o i n d i c a t e t h a t t h e s e r i e s on t h e right-hand s i d e i s an asymptotic expansion o f tion,

f(xj

x

as

tends t o i n f i n i t y i n the s ecto r

B

.

I n t h i s nota-

i s e i t h e r c l o s e d o r open.

;6

W e have d e f i n e d asymptotic expansions of f u n c t i o n s o n l y i n t h e case

x

when

tends t o i n f i n i t y .

I n t h i s c h a p t e r we s h a l l e x p l a i n v a r i o u s

p r o p e r t i e s of asymptotic expansions a l s o o n l y i n this c a s e . expansions o f f u n c t i o n s as

x

tends t o a p o in t

xo

Asymptotic

can be d e f i n e d and

d i s c u s s e d i n a similar manner without any e s s e n t i a l m o d i f i c a t i o n . 2.

I n t h i s s e c t i o n , we s h a l l prove

UniQueness o f asymptotic expansions.

t h e uniqueness o f an asymptotic expansion o f a f u n c t i o n as infinity

x

tends t o

. If

THEOREM 2.1:

m

( 2.1)

c

f(x)

cN"-N

N=o

and -

m

(2.2)

c

f(x)

-N yNx

N=O

then -

cN = yN

(2.3)

Proof:

(N=0,1,2,

...) .

We can assume without l o s s o f g e n e r a l i t y t h a t

g

i s a closed

sector.

Assumptions ( 2 . 1 ) and ( 2 . 2 ) mean t h a t , f o r every non-negative

integer

N

,

%

t h e r e are two p o s i t i v e numbers

I

f(x)

-

N C c ~ x 15 - ~ KNIxI

and

4 such t h a t

-N-1

k=o

and

for x E

for

xE

(0

.

Therefore

.

For

N=O

, we

get

co=yo

.

If we assume t h a t

c =y

P

P

for

PROPERTIES

p = 0,1,

..., N - 1

, then

and hence we g e t

3.

cN= yN

.

This proves Theorem 2 .l.

ea p ( x ) = Z cNx-N and -1 N=o The sum and t h e yNx-N be two formal power series i n x

P r o p e r t i e s o f a s n m t o t i c expansions. 0)

q(X)

= X

3

Let

.

N=o product of t h e s e two formal power s e r i e s a r e defined by 0

P ( x ) + q ( x ) = 2 (CN+yN)x-N N=o

(3.1) and

(3

=

.a

p(x)q(x)=

respectively.

N=o k=o

Furthermore, i f Pj(X) =

( 3-3)

N

x ( r.

m

c

CjNX

.

-N

(j = 1,2,. .)

N=o x -1 such t h a t

are formal power series i n c

( 3- 4 )

jN

for

=O

N=0,1,

* . * 9 n j'

and l i m n.

(3.5)

j + m

=a

,

J

then t h e sum o f t h e s e formal power s e r i e s w

m

x

c p (x) = x ( cjN)x-N N- j=i j=1 j

(3.6)

i s defined by

p j

m

.

m

Zc a r e sums of a f i n i t e number o f j =1 j N owing t o assumptions ( 3 . 4 ) and ( 3 . 5 ) .

Note t h a t t h e c o e f f i c i e n t s

Let us now consider a f u n c t i o n ables "hen

zk-%)

( zl,

...,zk) .

Assume t h a t

F( zl,

...,zk)

F i s holomorphic a t

(al,...,%)

.

.. , a , ) . ( zl-al, .. ,

(al,.

Set

(3.7) where

...+$ .

141 =tl+

NOW l e t us consider

k

jN

o f s e v e r a l complex v a r i -

F can be represented by a convergent power s e r i e s i n

i n a neighborhood of

c

formal power series

,

ASYMPTOTIC EDANSIONS

4

W

i R

x - ~

.

..

( j = 1, , ,k)

p.(x)=a.+ C c J N=l

(3.8)

By u t i l i z i n g t h e d e f i n i t i o n s given above, we can d e f i n e a formal

power series i n

x

-1

by

(3.9) We s h a l l denote by For example, i f

. . ., pk ( x ) )

F(pl(x),

p(x) = Z

t h e formal power series ( 3 . 9 ) .

i s a formal power series s a t i s f y i n g t h e

cNx-"

N=o condition

then

l/p(x)

denotes t h e formal power s e r i e s d e f i n e d by

Note t h a t

1/z = 1/[ co+( z-co) ] = co-1[ l+co -1( z-co) 3 -1

= c -1 z c-co-1( Z - C , ) ] N N=o 0)

(Z-C,)

N

N=o

F(zl,

Assume t h a t

...,zk )

i s holomorphic a t

(al,

...,%)

and t h a t

(3.11)

where j = 1 , ...,k

i s a c l o s e d s e c t o r d v e n by

B

8: a L a r g x L b

,

IxI2M

,

1x1 2

.

Let us c o n s i d e r a s e c t o r . . #

8: a l a r g x l b

Then, if (3.11) -

.

MI > 0 is s u f f i c i e n t l y larae, we have F(fl(x)

where t h e

Mt

p.(x) J

*

-

*

,fk(X))

'F ( P l ( x )

*

,Pk(X))

(X+

in

d .

0)

denote t h e formal power series on t h e right-hand sides of

PROPERTIES

Proof:

for

as

Since

xE

x

5

?i

if

MI > 0 i s s u f f i c i e n t l y l a r g e .

.

tends t o i n f i n i t y i n

Then (3.12) implies t h a t

Furthermore,

N -h f.(x)-a.= z c x + ~ ( x - ~ ) J h=l j h

as

x

tends t o i n f i n i t y i n

B

as

x

tends t o i n f i n i t y i n

B

I)

Assume t h a t

.

Therefore

-.

By u t i l i z i n g the d e f i n i t i o n of F ( p l ( x ) , . . . , p k ( x ) ) we can complete the proof o f Theorem 3.1. The following r e s u l t s a r e s p e c i a l cases of Theorem 3.1:

where

8

i s e i t h e r open o r closed.

11) Assume t h a t

Then

6

ASYMPTOTIC EXPANSIONS

6 i s a closeil s e c t o r

where

Let us consider a s e c t o r

Then, i f

cofO

M' i s s u f f i c i e n t l y large, we

and t h e p o s i t i v e number

have m

-N l / f ( x ) ~1," c cNx N=o

*

(x+m

~

j

LB) .

in

W e s h a l l now t u r n t o a j u s t i f i c a t i o n of term-by-term

i n t e g r a t i o n and

d i f f e r e R t i a t i o n of asymptotic expansions. T I I E O M 3.2:

Assume t h a t

3 .i3) where

B

i s e i t h e r open o r c l o s e d .

(3.14) where t h e p a t h of i n t e g r a t i o n should be taken i n Proof:

We assume without l o s s of g e n e r a l i t y t h a t

.

B rD

i s a closed s e c t o r .

Since

we have

hrthermore

sm[

N

x N

ch5-hld5=

h=2

h=2

c

h x-h+l -h+l

This proves Theorem 3 . 2 .

THEOW 3.3:

6

be a closed s e c t o r d e f i n e d by

rD: a L a r g x l b

where &

i;i

gi&

b

,

1x1

are real numbers and

t h e i n t e r i o r of

rD :

LM , M

i s a p o s i t i v e number.

Denote

7

PROPERTIES

fit:

,

a<arg x< b

1x1 > M

.

Assume t h a t f(x)

m

=

c

.I

&

(x+=

"$-A

0)

.

N=o m

-N-1 f l ( x ) 5 I: (-N)cNx N-

where Proof:

fl(x)

(x+m

denotes t n e d e r i v a t i v e of

f(x)

&

fit)

,

with respect t o

x

.

Let us put N

P,(X) =

z

ChX

-h

h=o

and = f ( x ) -pN(x) We must show t h a t

-N-2)

%(x) = O ( x

s u b s e c t o r of t h e open s e c t o r sector

9

.

gl

'

as

x

tends t o i n f i n i t y i n a closed

To do t h i s , l e t us c o n s i d e r a closed

defined by

where a<;<E
%>M.

Set

.,

If

M

8 = min(a"- a , b - b)

.

i s s u f f i c i e n t l y l a r g e , we can use t h e Cauchyfs i n t e g r a l r e p r e s e n t a -

ti on

(3.15) where

rx

i s a c i r c l e d e f i n e d by

rx: !%-XI

= 1x1 s i n 8

.

( S e e Fig. 3 . 1 )

ASYMPTOTIC EXPANSIONS

8

U %

\ \

\

\

0 Fig. 3.1. In f a c t , i f sector

&

&

f o r every x i n the closed s e c t o r on the c i r c l e Tx i f x E 9 , vhere

independent o f

is i n t h e closed -B . rxHence lEN(s)l 5

is s u f f i c i e n t l y l a r g e , t h e c i r c l e

x

.

Furthermore, we have

KN

i s a p o s i t i v e constant

151 2 (1- s i n 8 ) l x I

on t h e

-N-3

we can d e r i x e from t h e i n t e g r a l r e p r e s e n t a t i o n (3.15) t h e following e s t i mate: l%(x)l

This proves Theorem 3 . 3 .

sin

e )-N-1

(sin d l x l

-N-2

for

x

~

.i

9

PROPERTIES

One of the most important applications of asymptotic expansions i s

found i n the study of asymptotic evaluation of zeros of various functions. I n such a problem, t h e following theorem* is very u s e f u l .

LIBOREM 3 .&:

9"

Let two s e c t o r s B

be given by

61: a i a r g x l b

,

1x1 LM

Z:

,

1x1 2R

and z i a r g x
Assume t h a t

respectively. (i) f ( x )

i s holomorphic i n

(ii) f ( x )

tends t o zero as

(iii) a < a " < g < b ,

(iv) b - a l n

Then, if

61 ; x

B ;

tends t o i n f i n i t y i n

GLM;

. i s s u f f i c i e n t l y large, t h e mapping defined by y=x(l+f(x))

i s univalent i n Proof:

i.

I n the same way as w e did i n the proof of Theorem 3.3, we can

prove t h a t xfl(x)+O where

as x+w

p

and

line-segment

L

joining p

Note t h a t , i f

p

q

and

(3.16)

5

Idy/dx-lI and

q

are i n

q

on

L

51/3 and

g

in

P

1pI

.

Hence

y(p) = y ( q )

have

s

.

If a

(dy/dx)dx=O

L

IpI

.

and

lql

.

are sufficiently

Furthermore

,

and

follows from (3.16) t h a t

*

, we

is i n L

x

.

and t h a t

is i n

on

with respect t o

f ( x)

B , and

are i n

q

l a r g e , then every point

to

Z ,

in

are points i n

Suppose t h a t

p

x+w

f l ( x ) denotes the d e r i v a t i v e of dy/dx+l

if

as

For example, see T. Klmura [ 21; p .216].

lql

are s u f f i c i e n t l y l a r g e .

It

ASYMPTOTIC EXPANSIONS

10

Hence

.

p=q

i s s u f f i c i e n t l y l a r g e , t h e mapping

This means t h a t i f

y = x ( l t f(x))

i

i s univalent i n

.

4 , Summation of formal power s e r i e s .

I n many i n s t a n c e s of a p p l i c a t i o n of

asymptotic expansions, i t i s required t h a t we construct a function which admits a preassigned power s e r i e s a s i t s asymptotic expansion.

In this

The idea i s t o modify the given

s e c t i o n we s h a l l p r e s e n t such a method*.

formal s e r i e s i n such a way t h a t we o b t a i n a convergent s e r i e s by which a desired function i s d e f i n e d .

m:If Proof:

Re[z]
W e s h a l l begin with the following two lemmas.

, then

Il-eZ1
.

Note t h a t z

l-eZ=J

ecq ,

0

where the path of i n t e g r a t i o n i s t h e line-segment j o i n i n g assumption gration.

Re[z]< 0

implies

I ec I 5 1

Hence

Re[C]LO

f o r every

c

f o r every

6

z

to

0

.

on t h e path o f i n t e -

on the path of i n t e g r a t i o n .

This,

i n t u r n , implies t h a t

&!"A 1.2:

L e t a closed s e c t o r

g&

a

b

Re[eiep]<-6

.**

M

i s a p o s i t i v e number.

p,S, and a r e a l number

p

,

a p ( a r g xP(bp xE9

M ,

f o r every

For any p o s i t i v e number

f o r every

1x1 2

a r e real numbers and

t h e r e exist two Dositive numbers

Proof:

be given by

,

P: a ( a r g x ( b

where

B

, we IxpI

x

g

8

.

have

k#

Therefore, i f we s e t and

p = n/[ 2 ( b-a) ]

*+*See, f o r example, W. Wasow [47; As t o powers of x , w e use t h e

8 = n - p( a+b)/2

pp 40-43]. definition r x =exp[r(loglxl + i arg x ) ] for any complex constant r

.

The

,

8 such t h a t

SLTPATION OF

FORMAL POWER

SERIES

11

i t follows t h a t % = n - ~ ~ a r g [ e i ' x p ] < n +P

4

for

xEQ

.

I n other words, we have

for

xE &

.

Therefore, u t i l i z i n g the assumption:

6 > 0 so that

find

Re[ei8xP]<-6

for

x€B

o - -,+n 2

1212 @ > 0 , we

.

can

NOW we s h a l l prove the following theorem.

THEOREM L.1:

L e t a closed s e c t o r 8 61: a l a r g x L b

where

g&

a

b

,

be given by

1x1 2

a r e r e a l numbers and

M , M i s a p o s i t i v e number.

m,k t

a formal power s e r i e s

be given. where the f(x)

Proof: p(x) f(x)

cN a r e complex numbers.

which i s holomorphic i n

Then t h e r e exists a function

&

As i t was mentioned above, we s h a l l modify the given formal s e r i e s

s o a s t o obtain a convergent series by which the desired function w l l l be defined.

As t h e f i r s t s t e p , l e t us choose a sequence

(bN]

of p o s i t i v e numbers s o t h a t the s e r i e s

i s uniformly convergent f o r

1 XI 211 .

As the second s t e p , set

W

f (x) = 2

c ~ " - ~ [ exp(bNeiexP) l]

,

N=o where

8 and p are constants given by Lama 4.2. Re[vie2]<-%b<

0

for x € &

Observe t h a t

,

6 is a l s o a constant given by Lemma 4.2. Therefore, by v i r t u e of Lemma 4 . 1 , we have

where

I1-exp(bNeie~)I
for x E B

,

ASYMPTOTIC EXPANSIONS

12

'his means t h a t

f ( x ) is well defined and holomorphic i n

s t e p , we s h a l l prove (4.1). cN x - " ~

.

8

As t h e l a s t

To do this, n o t e t h a t

- exp(seiexP) 3

cNx-*

(x

N

, we

I t follows t h a t , f o r every p o s i t i v e i n t e g e r

N Z chx-h [ l - e x p ( b h e

-+= i n

.

s>

have

N

i* xP ) ] " = C

h=o

(x+w

ChX -h

in

8)

.

h=a

Furthermore, m

Z

chx-h [l-exp(bheie$)]=O(x

hd+1

No

Let

be a p o s i t i v e i n t e g e r such t h a t

f(x)=

-N-l+p

.

p
in

(x+-

8)

.

Then

N+No -h C chx [ l - e x p ( b heiexp)] h=a (D

- exp( bhsiep)

f r ,

]

h'N+No+1

=

N+No

z

ChX

-h

-N-No-l+p to(x )

(x+= i n

9)

.

h a Since

p-No
and

, we

N+No> N

have

N -N-1 f ( x ) = 2: c h x - h + ~ ( x h-

Tbe i n t e g e r

N

(x+-

in

Q)

.

being a r b i t r a r y , t h e asymptotic expansion (4.1) i s estab-

lished.

5.

h V m Dt o t i c

coefficients constants.

expansions o f functions dependinp on parameters. cN of t h e asymptotic expansion of

In many cases, however,

s = (el, sector, the notation

i l i a r y parameters

..., k ) . E

f

means t h a t t h e r e exist p o s i t i v e numbers that

f(x) have been complex

a s well a s

%

In the case when

(5.1)

(x+-

in

So f a r t h e

may depend on auxfi

i s a closed

o m i f o n d s for eEn)

% , independent

of

E

, such

13

FUNCTIONS DEPENDING ON PARAMFXERS

for

xEB

,

r E C l and

N)_O

.

If

B i s an open s e c t o r , notation (5.1)

means t h a t the asymptotic expansion i s uniform on each closed subsector of

B

,

Theorems corresponding t o Theorems 2.1, 3.1, 3.2 and 3.3 can be proved

a l s o i n this general case.

Now we s h a l l s t a t e a theorem which i s a refinement of Theorem 4.1.

TlJEoREM 5 . 1 : Let a closed s e c t o r B be Riven by

,

8: a l a r g x ( b

where a

gnJ

b

1x1 2

M ,

a r e r e a l numbers and M

i s a p o s i t i v e number.

&,

let

m

I: c ~ ( E ) x -be ~ Riven, where c N ( e ) a r e bounded

a formal power series

N=O

...,ck)

and holomomhic i n c = (el, t h e r e e x i s t s a function

i n a domain n & I s - mace. which i s holomorphic f o r x E & g&

f(x,c)

and -

a,

z

f(x,e)

cN(c)x-N

(x+-

in

uniformly for

cE 0 ,

c~n),

N=O

The proof of this theorem i s l e f t t o the readers.*

I n order t o s t a t e another theorem which is useful i n a p p l i c a t i o n s , we

f(x,c)

s h a l l assume t h a t

be holomorphic f o r

x E 8

,

161

5 6 , where

8

is a closed s e c t o r given by

,

8: a ( a r g x ( b

1x1 L M

,

t h e q u a n t i t y 6 i s a p o s i t i v e constant, and

101

that

where c N ( s ) a r e supposed t o be holomorphic f o r

=maxlej) j

181

(6

.

.

Assume a l s o

Let

and

be t h e Taylor's expansions of

*

See, f o r example,

f

and

cN r e s p e c t i v e l y i n t h e domain

K.O. F r i e d r i c h s [12; pp 152-153,BI.

14

ASYMPTOTIC EXPANSIONS

I €1 2 6 .

We s h a l l prove t h e following theorem.

THEOREN 5 -2: Under t h e assumptions s t a t e d above, w e have (5.2)

Proof:

.

(t1,...,$)

f o r e v e r y index Put

...,k )

c.(f3.) = p e x p [ i e . ] J

J

(j=1,

J

and

where

p

i s a f i x e d p o s i t i v e number.

.

p56

Assume t h a t

Then

and

2n C

M1.. .2,

=

...s2n

k

c,(e(G))exp[-i

0

Z &.0.]de, j=1 J J

0

... dok .

Since N

r ( X , E ( e )=) z h=o as

x tends t o i n f i n i t y i n

ch(e(e))x-h + o ( X - ~ - ' ) uniformly f o r

$J

...,k) ,

05e.52n J

(j=l,

we o b t a i n ( 5 . 2 ) .

DEF'INI TION 5.1:

xE

B,

c-sDace.

E

E

Let

coefficients g(x,4#O

for

Assume t h a t

, where

8

f(x,E)

g(x,c)

be holomorphic f o r

0 i s a domain i n

i s a s e c t o r i n x-plane and

m

C c ~ ( E ) x - be ~ a formal power series i n N=o cN(e) a r e holomorphic and bounded i n 0

.

x E B & €En . W

f(X,E)

Eg(x,c)

z

cN(")x-N

(x+-

in

x

-1

whose

Assume t h a t

8)

N-

uniformly with r e s p e c t t o

E

E means ~ that W

f(x,o)/g(x,s)

Z N=O

CN(E)X-N

(x+m for -

& uniformly

E E n).

CHAPTER 2 DEFINITION OF SUBDOMINANT SOLUTIONS

6.

An existence theorem.

(6

y"

0 1 )

Let us consider a d i f f e r e n t i a l equation

- P(x)y = 0

2

( y " = d y/dx

2

,

where

m-1 + a xm-2 t p ( x ) = x m t ax and

al,

...,am

1

2

...+am-1 x t a m

I n t h i s chapter we s h a l l define

a r e complex parameters.

.

subdominant s o l u t i o n s of the d i f f e r e n t i a l equation (6 .l)

The d e f i n i t i o n

i s based on a theorem expressing t h a t c e r t a i n solutions s a t i s f y i n g asympt o t i c conditions a t i n f i n i t y e x i s t .

Such s o l u t i o n s a r e characterized by

asymptotic conditions a t I n f i n i t y , and we must e s t a b l i s h t h e i r a n a l y t i c p r o p e r t i e s with respect t o the parameters

shall s t a t e such an existence theorem.

al,

...,am .

I n this s e c t i o n , we

We s h a l l a l s o present c e r t a i n I n Section 8,

important p r o p e r t i e s of subdominant s o l u t i o n s i n Section 7 .

we s h a l l i d e n t i f y subdominant s o l u t i o n s with known f'unctions when m = l and

2

.

mOR.EM 6 .L: The d i f f e r e n t i a l equation (6.1) has a s o l u t i o n

(6.3)

Y=l"(x,al,""am)

such t h a t ( i ) bm(x,a) i s an e n t i r e function of

(x,al,

...,am) ;

(ii) Qm(x,a) admits an asymptotic r e p r e s e n t a t i o n 7

(6.4) uniformly on each compact s e t i n

*

(al,. ..,a m )-space as

x

tends t o

This chapter i s mainly based on Po-Fang Hsieh and Y. Sibuya [17].

16

SUBDOMINANT SOLUTIONS

i n f i n i t y i n any closed s u b s e c t o r of t h e open s e c t o r

(6.5) where -

(6.6)

Em(x,a) =me2

rm

9

h,N and

then the Q u a n t i t i e s

'N'= l %,Nx

2

are polynomials i n

m,N

rm and A

.;I-' m

(al,

...,am) .

If we put

1

are given, r e s p e c t i v e l y , b-y

m,N

1

(6.8)

1 F( mt2-N)

L(me2) mtl

2

(m: odd) ,

1

-m - b

4

(m: even)

,

$1

and -

~ 1( m t 2 - 2 h ) i(mt.2-N)

mtl

'

(6.9)

=

N = l Am,Nx

1 l
of

b t h e n a o r e , the d e r i v a t i v e b;(x,a)

2 mt2-2h bhx

b m ( x , a ) with r e s p e c t t o

x

admits an asymptotic r e p r e s e n t a t i o n 1

1

( 6 .lo) uniformly on each compact set i n

.. , am )-space

(al,.

i t y i n any closed subsector of t h e open s e c t o r

...,

polynomials i n

as

x

tends t o i n f i n -

( 6 . 5 ) , where

.

are -

crn,N

(a1, am ) The only s i n g u l a r p o i n t o f t h e d i f f e r e n t i a l equation (6.1) i s

Furthermore,

P(x)

is linear i n

(6.1) i s an e n t i r e function of

.

( a l , , .,a )

e n t i r e functions o f

(al,.

Therefore a s o l u t i o n o f

...,

(x,al, am ) i f i t s i n i t i a l values a r e Since the s o l u t i o n b m ( x , a ) i s defined

.

m by the asymptotic condition (6.4) a t

I t must be proved.

.. , a m ) .

x

property ( i )i s n o t e v i d e n t .

= O J ,

The asymptotic r e p r e s e n t a t i o n (6.10) o f

derived from the asymptotic r e p r e s e n t a t i o n (6.4) of

manner.

l,jm

-r

m

exp[Em(x,a)b m ( x , a )

.

Then

+

F ' =[-r x -1 E;(x,a) t b ; ( x , a ) / t m ( x , a ) ] F

m

b;

can be

i n t h e following

Set

F =x

x=w ,

,

SUBDOMINANT SOLUTIONS

17

and hence [ F h m ( x , a ) h $ x , a ) = F ' t rmx-lF- EA(x,a)F Note t h a t (6.4) means

1

-P Bm,N~

W

Z

F"1+

.

N=l Therefore, by v i r t u e of Theorem 3 . 3 we have 7

-$NF f Z Z ( -N)Bm,Nx 1 -2 N=l QD

1

Hence

where

m,N

are

Thus we o b t a i n (6.10)

.

Set

P(x) = ( x - xl)

(6 .ii) where

kl,.

5 , ...,am

..,X m

( x - A,).

. . ( x - Am)

,

are r o o t s of the polynomial P(x)

are symmetric polynomials i n

.

It i s known that

. . . , I m ' Therefore t h e s o l u t i o n b m ( x , a ) i s an e n t i r e function of ( x , h l , . .,h ) which i s

(Al

symmetric with r e s p e c t t o B m,N

and

C m,N

A1,

,...,Am) .

.

The q u a n t i t i e s

a r e symmetric polynomials of

(Al,

m

rm

.. . , A m ) .

,

,

Am,N

I n the next s e c t i o n , we s h a l l d e f i n e subdominant s o l u t i o n s of (6.1) by

u t i l i z i n g Theorem 6.1.

We s h a l l a l s o p r e s e n t c e r t a i n important p r o p e r t i e s A proof o f Theorem 6 . 1 will be given i n Sections

o f subdominant s o l u t i o n s .

9 and 10. 7.

Subdominant s o l u t i o n s . ?=ei'x

where

(7.1)

Let us change t h e independent v a r i a b l e

by

,

8 i s a real number. d2y/&2

x

The d i f f e r e n t i a l equation ( 6 . 1 ) , then, becomes

- e-i(&2)'Q(%)y

=0

where

m

Am-h ~ ( 2 =P+ ) z ei m 8hx h=l

Therefore, i f we choose 0

so that

.

SUBDOMINAN T SOLUTIONS

18

the function

is

ii

This means t h a t , i f w e s e t

s o l u t i o n of ( 7 . 1 ) .

('?.zj

w=exp[i

2 R+2

bm,k!x,al

,...,am ) = bm ( w -k x,w -k a l , w

and

(7.3) k

where

is an i n t e g e r , t h e

-2k

a2

,...,w -mk a,) .

are s o l u t i o n s of ( 6 . 1 ) .

Q,,,(x,a)

In

particnlar

.

bm,,(x,a) =bm(x,a)

'fie following theorem i s a d i r e c t consequence of Theorem 6.1.

?a:

m O R E M

The s o l u t i o n

\a

satisfies t h e f o l l o w i n g conditions: m,k i s a n e n t i r e f u n c t i o n o f ( x , a l ,..., am) ;

(i! bm,,(x,a)

b

( i i ) bm,,(x,a)

I

m,k

(x,a)

admit asymptotic r e p r e s e n t a t i o n s

b,,,(x,a)~Ym(w

(7.4) y;,,(x,a)

-k su

-k

x,w

-k

-k

-k x,o

Y ; ( W

uniformly on each compact set i n

-2k

al,w

(al,

al,w

a2

,...,w -mk am) ,

-2k a2,

...,am )-space

...,w

as

-mk

x

am' tends t o infin-

i t y i n any closed s u b s e c t o r o f t h e open s e c t o r

l a r g x - - 2k m+2

(7.5)

where tively

& Y;

Ym

1'

3n

<m+2

9

are t h e right-hand members o f ( 6 . 4 )

.

Let

%

be t h e open s e c t o r d e f i n e d by

I arg

(7.6)

x

2k - mt2 -

TI

<m~2

-

and l e t

%

be t h e c l o s u r e of

(7.7) Sector (7.5) i s

2.k l a r g x-- m+2

s-l\

'I

V - Y %+1

9

Sk : TI

2-mt2

.

*

(See Fig. 7.1.)

(6.10) respec-

SUBDOMINANT SOLUTIONS

19

0 H g . 7.1. The m+2 s e c t o r s ( 7 .a)

-

, where

k=0,1,.

..,m-tl

(mod. m+Z)

cover x-plane completely.

DBINITI ON 7.1: zero as

x

I f a s o l u t i o n of t h e d i f f e r e n t i a l eauation (6.1) tends t o

%.

of t h e d i f f e r e n t i a l equation (6.1) tends t o i n f i n i t y as i t y along any d i r e c t i o n i n t h e s e c t o r

be dominant i n the s e c t o r

%

% , then

x

%

,then If a s o l u t i o n

tends t o i n f i n i t y a l o any ~ d i r e c t i o n i n the s e c t o r

t h i s s o l u t i o n i s s a i d t o be subdominant i n the s e c t o r

tends t o i n f i n -

this s o l u t i o n i s s a i d t o

.

Since

the solution Therefore,

b,

bm,k

i s subdominant i n

8,

and dominant i n

i s subdominant i n

%

and dominant i n

and

8-1

.

8,-1 and %+1

.

SUBDOMIN AN T SOLUTIONS

20

b m,o and b

The two s o l u t i o n s i s subdominant

bm,o m,l of (6.1) be subdominant i n go = ~

Y(X)

where

c

and

c1

subdominant i n

8,

0

a r e l i n e a r l y independent, s i n c e

, the

go

i s dominant i n

, and

.

Let a s o l u t i o n y ( x )

set

~ b C 1~b m , l, ~ + 9

a r e independent of

Y(X)

THEOREM 7.2:

bm,l

and

q u a n t i t y c1

=cobm,o9 co Ym

x

.

Since the s o l u t i o n

must be zero.

y(x)

is

Hence

.

The s o l u t i o n bm(x,a) i s t h e only s o l u t i o n of (6.1) which

s a t i s f i e s all of the resuirements l i s t e d i n Theorem 6.1.

Urn

The uniqueness of

.

implies a s i m i l a r uniqueness of

m,k As an application of the uniqueness of b m ' we s h a l l prove another theorem which we s h a l l u t i l i z e a t various places i n the following chapters. THEOREM 7.3 :*

Let x,

,...,am

and X be complex v a r i a b l e s . the function bm(x,a) s a t i s f i e s the i d e n t i t y

(7.9)

s , a1

bm(x+s,al,.

..,a m-1'

= Km(x,al,. where Km, -

.

ul,. .,u

respectively,

&

m

am+X)

..,a m )bm (s,ul,. ..,"m-l,um+X)

a r e independent of

{'

and (x+s)m+al(x+s)m-l+

m m-1 = s +up Aa a function o f

Proof:

9

and a r e defined,

, even) ,

(m: odd)

exp[-Em(x,a)]

(7.11)

X

s

r

Km =

( 7 .lo)

Then

s

,

(m:

...+ am-l(x+s) + am

+ '..+um-ls+um

l,jm(x+s,al,

.

...,am-1' am+A)

s a t i s f i e s the

d i f f e r e n t i a l equation d2y/ds2

(7.12)

- [ sm+ulsm-1 +

...+

s + u m +Xly = 0

and lim y = 0

( 7 -13)

S++=

where

lim

mean8 t h a t t h e l i m i t i s taken a s

S-+*

*

See Y . Sibuya [34; p 561.

s

tends t o i n f i n i t y al ong

SUBDOMINANT SOLUTIONS

t h e p o s i t i v e r e a l a x i s i n s-plane. of

bm(x,a)

, we

(7J 4 )

Therefore, by v i r t u e of t h e uniqueness

have

..,a m-1' am+A)

bm(x+s,al,.

= Km bm ( s , u ly...,u m-1' um+X) Km

where

21

i s independent o f

s

.

,

.

W e s h a l l compute Km

F r o m (6.4) and (7.14) we d e r i v e

r

+

( x + c ~ )1~ [O((x-ts)

--12

) ]exp[ -Em(x+s ,al, 1

--

F

...,a m-1'

am+A)

...,um-1' um+A)] r i s a polynomial i n (al, ...,am-1' m (ul,...,um-1' um+ X ) . If m i s odd,

3

= Kms m [ l + O ( s 2 )]exp[-E,(s,u1, as

s tends t o

+m

, where

?m i s a polynomial i n have r = F merefore m and

m--r.

Km= l i m

am+)i) we

+ Em(s,uA)1 ,

exp[-Em(*s,aA)

S-)*

where

aX = ( a l

,...,am-1'

am+X)

(6.9) implies A m , N = O

^A

where

m,N

if

and

N i s odd,

a r e polynomials i n

,...,um-1'

\=(ul

Therefore a s

(x,al,

-

l(ntt-2) m+l

Em(s,uh) where/

Since

m,N

Km

are polynomials i n

..,m+l)

when m

i s odd. m

, and

hence

$,Ns

N=l

...

I$,= 1

.

1

,

-r

m

, we

$(m+2-N)

( u ~ , , um-1' um +X)

i s even, we have r = -m-b m -4

where

+

+-

Set

9

.

Then

i s f i n i t e and d i f f e r e n t f r o m zero, we must have

( N =1,2,.

If

=*2~

Observe t h a t

tends t o

s

...,am-l,am+X) .

2 2

.

um+ A )

-

n

*m,N = %,N This proves Theorem 7 . 3 f o r the case

1

-

=--m-b

4

-

1 m-1 [1+ I: anx-"+ ( a +X)x-m ]2 =1+ WI: bhx-h m h=l n=l

22

SUBDOMINANT SOLUTIONS

and

m-1 [lt

z

uns

-n

-I

m

-m 2 t (um+A) s ] = 1+ Z bhs-h

n=l Since

k,N=Oi f

N

h=l

i s odd, w e have l ( m t 2 ) mil 2

Emk+s,%) = z s where

m,N

a r e polynomials i n

' N'= l Am,Ns

2

(x,al,

...,am-1'

$mt2-N) 1

+E,(X,+

.

a ti)

m

Hence

m Kmexp[Em(x,y)]= l i m exp[ Z s++ N=l t

(bl

I t i s not d i f f i c u l t t o prove t h a t

8.

Case m = 1

and case

Ai(x) =

(8.1)

m = 2

1 2ni

~ 1( w 2 - N )

(i m,N -^A m,N )s

-bl

F1

)log s ] = 1

.

Zm+l

Em(x,ak)= E m ( x , a )

.

.

.

I t i s known t h a t Airy's i n t e g r a l

exp(xt-t3/3)dt

s a t i s f i e s t h e d i f f e r e n t i a l equation y" - q r = 0

(8.2) where

r

is t h e path of i n t e g r a t i o n which i s given by F i g . 8.1.4

Fig. 8.1.

*

See, f o r example, W.Wasow [ 47; pp 124-126 3.

9

CASE m = l

23

ANDCASE m=2

Equation (8.2) i s a s p e c i a l case of equation (6.1), where al=O

.

I n case

m=l

, we

w=exp[i

have 2

m+l=2

,

m=l

and

m + 2 = 3 and

.

n]=exp[i2n/3]

Therefore t h r e e s e c t o r s

-

s-~:

Iw g

X+

2d3i 9 / 3

so : lw XI 5 d 3 81 : I a r g x - 2n/31 2 n/3

,

9

cover t h e x-plane.

(See Fig. 8.2.)

Fig. 8.2.

The t h r e e s o l u t i o n s l$1,-1

’

a r e subdominant i n Sel , Q1,O and 1$1,1 Also, i t i s known t h a t Airy’s i n t e g r a l (8.1)

So , and S1 , r e s p e c t i v e l y . admits an asymptotic representation*

as

x

tends t o i n f i n i t y i n

Ch

h = (-1)

8-1vgov81

eP

where

5 4 ~ ~ ~ 1~ h ~’ - - )

.

i s subdominant i n 8, Since a i ( x + a , ) satisf i e s equation (6.1) f o r m = l , t h e uniqueness of implies t h a t This m e a n s t h a t

Ai(x)

(8.4)

bl(x,al) = A A i ( x + y)

*

urn

.

See, f o r example, M. Abramowitz and I . A .

S t e p [l; p448, 10.4.591.

2L

SUBDOMINANT SOLUTIONS

I t f o l l o w s from ( 8 . 4 ) t h a t

I

,

'11,-1(x,a,) = W;;lli(u(x+ a l ) )

b

(8.5)

190 1 9 1

(x,a,) =Z&Ai(x+al)

,

(x,a,) =2J;;Ai(w-'(xt

a,))

.

I d e n t i t y ( 7 . 9 ) means, i n this c a s e , t h a t Ai(xt s+ al+h) = ~ i ( s + u ~ + A ,) where

a1 = x t a 1 ' I t i s a l s o known t h a t t h e i n t e g r a l

1

r(2-a) TJ(a,x) =

(8.6)

2ni

1

1 2

e

'

a-2 t

7

1 2 exp(xt-Tt )dt

C

s a t i s f i e s t h e ? i f f e r e n t i a l equation y"-

(8.7) where

1 2

(7t

4

a)y=O

,

C is t h e p a t h o f i n t e g r a t i o n which i s given by F'ig. 8.3.*

F'ig. 8.3. A s i s w e l l known, i n t e g r a l (8.6) admits an asymptotic r e p r e s e n t a t i o n

u n i f o r d y on each compact set i n a-plane as c l o s e d s u b s e c t o r o f t h e open s e c t o r :

larg

x

XI

tends t o i n f i n i t y I n any

< 3n/4 .**

I t is e a s i l y

verified that

* %*

See, f o r example, M. Abranowitz and I . A . See I b i d . , p 689, 19.8.1.

Stegun [l; p 687, 19.5.11.

CASE m = l

1 1 2 y = u(Fa2 - ~5,4

(8.9)

AND

x+

.

r: ml-2

25

1 F ~) )

s a t i s f i e s equation (6.1) for m = 2 1 I n case m = 2 , we have ~ + = l 2 w=exp[i

CASE m = 2

,

m+2=4

and

n]=i

Therefore, f o u r s e c t o r s

cover x-plane.

(See Fig. 8.4.)

Flg. 8.4.

The four s o l u t i o n s I$

S-l

,

8,

,

81

,

'

2,-1 and 82

,

.

a r e subdominant i n

U(a,x) implies t h a t s o l u t i o n (8.9) i s subTherefore, by v i r t u e of t h e uniqueness of ,$l , w e have

t a t i o n (8.8) o f t h e i n t e g r a l dominant i n So

' h,l '

and b , 2 r e s p e c t i v e l y . Also, t h e asymptotic represen-

b,o

(8 .lo) It follows from (8.10) t h a t

SUBDOMINANT SOLUTIONS

26

If w e denote

by

D,,(x)

t h e Whittaker's p a r a b o l i c c y l i n d e r f u n c t i o n , we

have U(a,x) = D

,(x) -a--

.*

2

I d e n t i t y ( 7 . 9 ) , i n this c a s e , can be derived from t h e following identities: 2

u =x + a x t a 2 1 2 '

u1=2x+al,

~ ( , fci ( x + s + F1 al)) where

=U(C

1 1 2 1 1 2 2=2-zal=zu2-P1

,Ji(s+;;ul)) 1

.

i s an a r b i t r a r y complex q u a n t i t y .

c

To the knowledge of t h e author, t h e r e are no well-known functions

with which

9.

can be i d e n t i f i e d f o r

Proof o f Theorem 6.1.

( 9 .I)

u'=A(x)u

m23

.

Equation (6.1) i s equivalent t o t h e system

,

where

u =

[j

,

A(x)

By the change of v a r i a b l e s u = [ 10 system ( 9 -1) i s reduced t o ~

*

~~

See, f o r example, M. Abramowitz and I . A . S t e p [l; p 687, 19.3.71.

'

PROOF OF THEOREM 6.1

27

where

3

A0 = [ and

(al,

pN

9

a r e two-by-two matrices whose components a r e l i n e a r i n

...,am ) .

Let us s e t

t o obtain dw/d5

where components o f ( a y . , am1

9

Let

p

and

we have

Hence, i f we s e t

we get

B(5)

y

=

5-l

a r e polynomials i n

and

-B ( 5 )

,

=S*’B(S)w

[i

I]

+ O(5-l)

and l i n e a r i n

.

be unknown q u a n t i t i e s , and l e t us s e t

28

SUBDOMINAN T SOLUTIONS

i

(9.2)

Y(S) = + ) +

,

B1WP

dP/@ = Smf1[

e2( 5 ) + a2 ( 5 )P - y ( 5 ) p I .

Equations ( 9 . 2 ) a r e e q u i v a l e n t t o

(9.3) W e s h a l l determine p

by t h e second e q u a t i o n of

be given by t h e f i r s t equation o f (9.3).

(9.3) and then y

In o r d e r t o determine

p

will

, we

s h a l l use the following lemma.

9.1:

f

are linear i n

,

h

(al,,..,a

g

m)

.

be polynomials i n

5-l

whose c o e f f i c i e n t s

Suppose t h a t

(9.4)

as

5

tends t o i n f i n i t y , where

( 5 ,a1,

pendent of

(9.5)

...,am) .

dp/dS

ho

i s a non-zero c o n s t a n t which i s inde-

Then t h e d i f f e r e n t i a l e q u a t i o n

=5&’[f(5)

h ( 5 ) p + d5)P21

-k

has a unique formal s o l u t i o n 9)

= 2 pNC-N

(9.4)

7

N =1

where t h e Q u a n t i t i e s pN are Do1.ynomiaI.s i n E m a t i o n ( 9 . 5 ) admits a s o l u t i o n

(al,.

p(5)

..,am )

.

s a t i s f ~ n pt h e f o l l o w i n g con-

d itions:

(ii

For each p o s i t i v e c o n s t a n t

constant

6

t h e r e e x i s t s a p o s i t i v e number

holomomhic with r e s p e c t t o

where r

6

(S,al,.

. .,am)

N

ri 6

such t h a t

p(5)

is

for

t a k e a l l v a l u e s i n a domain 0<6<6,,

6,

and each s u f f i c i e n t l y small p o s i t i v e

r

O < r < b ,

being a small p o s i t i v e number;

(ii) p ( 5 )

admits t h e f o r m a l s o l u t i o n

uniformly on each compact s e t i n

(al,

fj(S)

...,%)- space as 5 t e n d s t o i n f i n

i t y i n any c l o s e d s u b s e c t o r of t h e open s e c t o r

( 9-8)

Iarg

h o + (m+2)arg

a s i t s asymptotic expansion

51 < $

.

6.1

PROOF OF THEOREM

We s h a l l prove Lemma 9.1 i n Section 1 0 .

29

In this s e c t i o n we s h a l l

To do t h i s , we s h a l l apply Lemma 9.1

prove Theorem 6.1 by using Lemma 9.1. t o t h e second equation of ( 9 . 3 ) , where

f(5) = B2(9

?

By v i r t u e of the d e f i n i t i o n of

~ ~ ( = 5o(5-l) )

- a1(5)

h(S) =a,W

p,, and B,

al, a2,

,

~ ~ ( =0(5-~) 5 )

,

g(s) = -B,(5)

5

, we a2(5)

Therefore a l l requirements i n Lemma 9.1 are s a t i s f i e d . h =4 0

.

-

have

- a l ( < ) = A + o(s-') In p a r t i c u l a r ,

!&is means t h a t t h e second equation o f (9.3) admits a s o l u t i o n

p( 5 ) s a t i s f y i n g the following conditions: ( i ) For each p o s i t i v e constant r and each s u f f i c i e n t l y small p o s i t i v e constant

t h e r e e x i s t s a p o s i t i v e number

6

holomorphic with r e s p e c t t o

...,am )

(S,al,

Nr,6

such t h a t

p(5)

is

i n t h e domain

(9.7')

r

where

and

take a l l values i n t h e domain

6

(ii) we have m

uniformly on each compact s e t i n

..,a m )-space

(al,.

as

5 tends t o i n f i n -

i t y i n any closed subsector of t h e open s e c t o r

where

3n

I =g

(9.81)

51 < 2(&2)

pN are polynomials i n

3

(al,.

..,a m ) .

Set

(9 .lo) Then y ( 5 )

y(5)

=.,(sf

-F B1(5)P(S)

i s holomorphic i n domain (9.7 1 Q)

(9.11)

y ( s ) z-2+ 2 YNS N=l

uniformly on each compact set i n

)

, and

w e have

-N (al,

...,am )-space

as

tends t o

i n f i n i t y i n any closed subsector of t h e open s e c t o r ( 9 . 8 ' ) , where t h e a r e polynomiah i n

(al,.

.., am ) .

yN

.

SUBDOMINANT SOLUTIONS

30

Let

and

Then s e t (9.12)

and

If we c a ~ prove t h a t

u ( x ) i s an e n t i r e function of (x,al, t h e proo;' of Theorem 6.1 w i l l be completed. To do this, l e t

...,am )

, then

@ ( x ) be t h e

two-by-two matrix such t h a t 0 1 ( x ) = A ( x ) 4 ( x ), 0 ( O ) = l 2 where

l2 i s the two-by-two i d e n t i t y m a t r i x .

I t i s known t h a t

@ ( x ) and

.. .,a m ) , and u ( x ) =I ( X I 4 (xo)-%xo) .

are entire i n

#(x)-l

Let

,

0

(al,.

. . , am ) 0

(x,al,

be f i x e d but a r b i t r a r y , and consider a small neighbor-

...,

of this p o i n t i n t h e (a1, a )-space. Choose xo so t h a t the rn 1 2 is i n domain ( 9 . 7 ' ) f o r every ( a l , . . . , a m ) € V p o i n t (xo,al,. . , a m/ This means t h a t u ( x o ) i s holomorphic i n V Hence u ( x ) i s e n t i r e i n

hood

V

-

.

'

.

.

m1 * Let us now compute

h a 1 , ...,a

r

m

E(S)

and t h e d e f i n i t i o n o f

and A m,N we d e r i v e

i n ( 6 . 4 ) and (6.6).

1

m' = ?m+2

(9 -14)

i(mt2) m t l

yN

- 2 mt2-N N=l

A d i r e c t computation shows t h a t

1 z(mt2-N)

From (9.13)

PROOF OF THEOREM 6.1

31

(9.15)

Hence

-1

= -2[X^mp(X)] 2 -?S1 -m-2+ o(5-m-3)

-

set

.L

aD

[x-%(x)] 2 =1+. 2 bhx-h ,

h=l

This i s the same as ( 6 . 7 ) of “Heorem 6.1. Then we get c

I

for N = 1 , 2

(N: odd)

0

,

,...,Wl , and 1 m

(m: o d d )

(m: even)

,

.

SUBDOMINANT SOLUTIONS

32

This proves ( 6 . 8 ) o f Theorem 6 .l. We can prove ( 6 . 9 ) i n a s i m i l a r manner. Notice t h a t (6.6) o f Theorem 6.1 can be w r i t t e n as 1 Em(x,a) =-mt2

I n parti-cular

bhT

Z

-1

--1

-h

d7

.

-1

=x 4~1o +( x-2 ) 1 e q [ -273/2 - a1x2]

b,(x,al) and

_I

3

_I

I

i

--1

-

1

L

---( --aa ) 2 2 2 8 1 b2(x,y,a2) = x

These r e s u l t s agree with those i n Section 8.

.

Let us compute B m,N

To do this, set

X

(9.16)

Y = z eW[J p(T)dTl

7

0

where

1 I

p ( x ) =-xTpl[l+

bhx-h]

Z

.

l(h
2

t a ( x ) 2 I t ['p 1 ( x ) t ' p ( x )

I t i s easy t o see t h a t

- P(x) 12 = 0

.

,

A

rq(x) = x q -2+ O(x-1) ] 2

p'(x) + p ( x )

,

$4

- P(x) = x

[ z r m tO(x-1) 3

.

Hence we can f i n d a unique formal s o l u t i o n 1 r OD m z = x [1+ Z B m,? 1' N=l

-2

where the

Bm,N

a r e polynomials i n

(al,

...,am ) .

in order t o complete t h e proof of Theorem 6.1, we s h a l l prove Lema

9.1 i n the next, s e c t i o n . 10.

Proof of Lemma 9.1.

As t h e f i r s t s t e p of t h e proof of Lemma 9.1, w e

s h a l l construct a formal s o l u t i o n (9.6) of equation ( 9 . 5 ) . write (9.5) as S4-ldp/d5

=f(S)

+ h ( 5 ) p + g(S)p2

.

To do t h i s ,

PROOF OF LFNMA 9.1

33

Upon s u b s t i t u t i n g t h e formal series (9.6) i n t o ( 9 . 5 ) , we o b t a i n hoPN=fPN(P1,"'9PN-l;

%

a19"*Ya

m)

...,

.,.,

(N21)

,

.

This proves the pN-l; al, am ) (pl, e x i s t e n c e of a unique formal s o l u t i o n , and i t s c o e f f i c i e n t s pN a r e eviwhere

i s a polynomial i n

d e n t l y polynomials i n

( al,

...,am) .

Let s=- ho

ml-2 mt25

(10.1)

*

I n s-plane, consider a c i r c l e

1.

(10.2)

=M

and two p o i n t s

6 i s a small p o s i t i v e constant and M i s a posi-

on c i r c l e (10.2), where t i v e constant.

s =sl

and

$- 6

at

s=st

2 . sector:

Let

s = s2

s=si

T1

and

be t h e tangents t o c i r c l e (10.2) a t

, r e s p e c t i v e l y . Let T1 i n t e r s e c t t h e l i n e : , and l e t T2 i n t e r e e c t t h e line: a r g s = 6 - $ 8

L e t us denote by

(10.4)

T2

lug

91

6 tM

arg s = at

t h e Riemann domain which contains t h e

<3-6, Isl>M1

f o r a s u f f i c i e n t l y l a r g e p o s i t i v e constant

and which i s bounded by the

MI

following arcs:

(10.5)

(See Fig. 10.1.)

A t each p o i n t

s

of

8

6 ,M

we can c o n s t r u c t a s t r a i g h t

line

(10.6)

ie

a=s+re

(O
s o t h a t U n e (10.6) i s contained i n 86,M, and

SUBDOMINANT SOLUTIONS

34

U

Is1 = M

-7r

S

\

/s;

Fig. 10.1.

PROOF OF LEMMA 9.1

let 1 1p - 6 ) .

(10.7)

LEMMA 1.0.1: constants

(See F i g . 10.1.)

Let p o s i t i v e constants

s u f f i c i e n t l y small, while

M6 9 P

and -

Lo

p

35

6

p

is arbitrary.

be dvem. where

6

Then t h e r e exist p o s i t i v e

such t h a t

0

(10.8)

sE8 6,M6,p where

Lo

i s independent of

s t r a i g h t l i n e (10.6) -*Proof-

’

I n case

1.1

, and

p

.

2 1.

t h e path of i n t e g r a t i o n is t h e

f o r every p o i n t on t h e s t r a i g h t l i n e (10.6) i t

i s easy t o prove (10.8).

In f a c t we can s e t

.

L0 = [ s i n ( $ ) l - l

(See Fig. 10.2.)

10.2. Let us consider (10.8) under t h (10.10)

1. >

min

( s + wie

assumption tha

I

.

KT<+

(See Fig. 10.3.)

ie

= s+zp, I I I

LO Flg. 10.3.

36

Let

SUBDOMINANT SOLUTICPJS

so = s t Toeie

derive

T ~ 0>

.

I sol =

be such t h a t

min

Is+

i0 I

78

.

From (10.10) w e

%T
If w e s e t

we g e t

where

.

p=7-Iso-s(

Set

.

po=-Iso-sl

Then

2

ls12=D2+po

,

7=C1-p0

9

and

Set

1

Y(t) =(D2t t Then

and

Choose

M

6,P

s o large t h a t

2

TP t ) e

1 cose

J+-=(D2+C1 2 ) -2pe-p c o s 0 dC1 t

-

PROOF OF LEMMA 9.1

since

D> M

sE8

for

6,P

69M6,p

a

1

cos 8 2 s i n ( p )

We a l s o have

37

.

Hence

cos

e t p->- t

1 s i n ( 716 )

.

D2tt2 Therefore dY( t ) / d t and we obtain

> [$ sin($) 1

y(w0)

]Y(t)

1

< ep 0 s i n ( 2 6 ) [ Y ( o )

t

-1,

(3

1

1

-3 s i n ( 2 6 )

sin($))-'[e

2 0

- 111

.

< 3 [ sin($) 1-l

This completes the proof of Lemma 10.1. W e s h a l l now complete t h e proof of Lamma 9.1 by using Lemma 10.1.

do t h i s , consider a s e c t o r 8 i n the

( al,

...,a m )-space

? - p l a n e and a domain

To

;Qr i n the

which a r e defined r e s p e c t i v e l y by 8: l a r g ho+ (mt2)arg

53 ,

51

151 2n

and pr:

lall t

...+(%I


,

R i s a fixed p o s i t i v e constant. I t i s known t h a t t h e r e e x i s t s a function fir( 5) s a t i s f y i n g t h e following conditions:*

where (i)

cr(5) i s holomorphic i n (?,al

E ;Qr i (ii) fir(%) zfi(5)

,

,...,am )

d$r(S)/d?zdP(5)/dS

for

(5+=

SE8

and

(al

,...,am )

8 uniformly i n

in

Set P=qfPr(S) Then (10.11)

ddd?

=?mt1[wr(5)

Lbr(5)

=f(<)t h ( 5 ) $ r ( s ) t

+ h r ( 5 ) q + wr(5)q 2

I

9

where

Xr(S) = h ( ? ) + 2g(S)Cr(5) wr(?)

+%

=ids)

ds)br(5)2-5-m-1dbr(5)/d5 ,

a

See Theorems 4.1, 5.1 and 3 . 3 of Chapter 1.

9

;Qr).

SUBDOMINANT SOLUTIONS

38

W e have pJ5)

(10.12)

20

9

,

hr(S) =hot0(5-1)

$1

Br)

.

i s due t o t h e f a c t t h a t t h e asymptotic

(S+W

q(5) GO 8'

uniformly i n

r( 5 ) i s a formal s o l u t i o n o f (9.5). p

W e s h a l l construct a solution

where

8

,

=0(5-3

The asymptotic property of expansion $(s) o f $ , ( 5 )

(10.13)

in

(5+m

q

8'

in

denotes t h e i n t e r i o r o f

of (10.11) such t h a t

8

Br)

uniformly i n

.

,

To do this, we s h a l l consider an

i n t e g r a l equation

where

s

=-&ho mt2 25

Y

ho p o=-mt2

Then t h e i n t e g r a l equation (10 .I.,$)

2

becomes

S

(10.15)

q(s)

=$ J O

2

[ p , ( ~ ) -+ i r ( ? I ) q ( 7 )+ v r ( s ) q ( q )

le

s-u

do

.

W

W e s h a l l consider the i n t e g r a l equation (10.15) i n a domain (10.16)

s€

where constants

r

(a1,

9

and

6

...,a

Ear

y

a r e f i x e d , and t h e constant

M

w i l l be speci-

M must s a t i s f y t h e condition: M 2 I ho(fl%m+2). The path o f i n t e g r a t i o n is given by t h e s t r a i g h t l i n e (10.6). By v i r t u e of t h e asymptotic p r o p e r t i e s (10.12) of p r ( 5 ) , k r ( 5 )

fied l a t e r .

and

vr(c)

The constant

, we

have

1wr(5)1 5 L 1 5 r 2 ,


iL15r1

L is a p o s i t i v e c o n s t a n t .

i n domain (10.16), where lw(<)[

l:rwl

9

Ivr(5)1 IL151-1 Assume t h a t

i n (10.16), and s e t

~ ( 5 =h ) l S[pr(7) t X",(T)w(q) + ~ y ( l ] ) w ( ' ? l ) ~ ~. e ~ ~ d ~ o

Then

w

PROOF OF LEMMA 9.1

39

where

' L lhol

,

C =-[2+1/MI] O

Then

L L(1+2/M1)

Iv ( s 11 I +li i n (10.16), where [lwl-w.J

w1-w211 denotes

sup(wl(5)-w2(5)1

i n (10.16).

Thus, we can prove t h e existence of a s o l u t i o n ( q ( s ) (<151-'

i n (10.16).

I n o r d e r t o prove t h a t

q(5) q(5)

asymptotic condition (10.13), l e t us consider a function

such t h a t s a t i s f i e s the

w(5)

such t h a t

I w ( 5 ) 1 C_%151-N i n (10.16), and set

From the asymptotic property (10.12) of i n (10.16), where

4 i s a positive

-N-1

~ ~ ( ,5 we) g e t ( p r ( 5 ) 1 ihI51

constant.

Then

-N-1 Iv(5)I ~ ~ 1 1 5 1 i n (10.16) by using

Therefore we can prove an estimate Lemma 10.1.

Set

P(5) = q ( 5 ) + " p ( 5 )

(10.17)

This i s a s o l u t i o n o f (9.5).

*

If s o l u t i o n (10.17) i s independent of

r

and

6 , then t h e proof of Lemma 9.1 will be completed. be a s o l u t i o n of (9.5) which s a t i s f i e s a l l requirements i n the statement

To see this, l e t

of Lemma 9.1 i n the domain

(10.18) and l e t

p2(5)

be such a s o l u t i o n i n t h e domain

p1(5)

40

SUBDOMINANT SOLUTIONS

(10.19) If w e choose

6

,

i n a s u i t a b l e way, t h e domain

.

’

sE86,M

(10.20)

2

r

and

M

(al,

’

‘“,PI2

(a1,. .,am) E Br

i s c o n t d n e d i n domains (10.18) and (10.19).

4 5 ) =p,(S) -Pz(S)

Set

*

Then

=Srn+lJ(5)u(5) ,

du(!)/dS where

J ( 5 ) = h ( S ) + g(S)[pl(!) + p2(5) 1 u(5)

We s h a l l i n v e s t i g a t e

i n domain (10.20).

. Since

pl(S)

and

p2(5)

s a t i s f y the same asymptotic condition i n (10.20), we have

u(5)

(10.21)

ZO

s ~ uniformly , ~

in

(s+-

in

.

B ~ )

We a l s o have

J ( 5 ) =ho+0(5-’) Let

so

be an a r b i t r a r y p o i n t of

(s+=

in

s6,M, and

‘6,M

Br)

unifody in

.

set

Then

where

1

o ~ ~ , i l te b 7 , and for

s-6,M If

u(S,)#O

,

then

u(5)

=

0

.

tends t o i n f i n i t y as

i s a contradiction, since u

Thus wa g e t u(5)

t20

t

tends t o

+=

.

This

s a t i s f i e s t h e asymptotic condition ( 1 0 . 2 1 ) .

i n (10.20).

This completes t h e proof of Lemma 10.1.

PROOF OF

Remark:

0

LEMMA 9 -1

Theorem 6.1 has been generalized t o the d i f f e r e n t i a l equation: 2

x yfl-p(x)y=o

,

m

P ( x ) = x +alx

by F.E. Mullin [25; Theorem I on p.53,

m-1

-t . . . + a m

CHAPTER 3 Idwl

A RELATION BETWEEN Qm

11. Remarks on p e r t u r b a t i o n s of asymptotic s o l u t i o n s ,

*

Let us consider a

second order d i f f e r e n t i a l equation o f t h e form (11.1)

+ EQ(x)I y = o

y " - [P(x)

9

.

Q(x) a r e polynomials i n x Let us denote by p q t h e degrees of P(x) and Q ( x ) with r e s p e c t t o x , and s e t where

P(x)

Urn x"P(x)

and

= Po

,

lim x - ~ Q ( x ) =Qo

, and

k=max[p,q]

.

If we s e t

X-r-

X-bo

equation (11.1) i s equivalent t o u'=A(x,c)u

(11.2) Let us change

x

and

u

.

by

Then system (11.2) becomes

(11.3)

dv/d5 = tktlB( 5 , c ) v

,

where

*

The main p a r t of t h i s chapter i s based on F.E. Mullin [ 2 5 ] .

and

REMARKS ON PERTURBATIONS

6(c)

(11.5)

if P L 4

Po+O(d

=

p
if

CQO

9

.

!Ibis shows t h a t t h e leading c o e f f i c i e n t matrix o f

[

(11.6)

p< q

0 if

B(5,e) :

:]

6Fo)

has two d i s t i n c t eigenvalues & eigenvalue

43

.

<

if

p2q

, but

i t has a multiple

L e t us now consider a system of t h e general form

k

(11.7)

,

v l = x B(x,s)v

k i s a non-negative i n t e g e r , x i s a complex independent v a r i a b l e , c i s a complex parameter, v i s a n n-dimensional vector, and B is an n-by-n matrix. Assume t h a t t h e components o f B(x,c) are convergent power where

series i n c=O

x-l

, and

whose c o e f f i c i e n t s a r e holomorphic i n a neighborhood of

set

(11.8) a r e n-by-n matrices whose components are holomorphic i n where t h e %(e) a neighborhood of c = O I n t h e case when t h e leading c o e f f i c i e n t matrix

.

Bo(0)

matrix

has

n mutually d i s t i n c t eigenvalues, t h e r e exists an n-by-n

P(x,E)

(i) P(x,c)

such t h a t

and

are convergent power series i n x

P(x,E)-’

c o e f f i c i e n t s holomorphic i n

c

at

-1

with

c=O ;

(ii)’A(x,c) = P ( ~ , o ) - ~ B ( x , r ) P ( x , t ) i s diagonal..

The diagonal components of column vectors of unknown q u a n t i t y

(11.9)

P(x,c) v WI

A(x,t)

a r e eigenvalues of

are eigenvectors of

by v = P ( x , e ) w

k = [ x A(x,c)

, WB

and t h e

B(x,E)

.

B(x,c)

By changing the

obtain

- P ( x , ~ ) - ~ d P ( x , c ) / d x ] w.

Fix any d i r e c t i o n .arg x = 8

.

Then i n the x-plane, and set x = teie P(x,C)-ldP(x,e)/dx is i n t e g r a b l e a s t tends t o t = Therefore, a s f a r a s t h e asymptotic behavior o f s o l u t i o n s of system (11.9) ( a s t tends t o

.

ia

) i s concerned, t h e system

(11.lo)

k z l = x A(x,t)z

w i l l give a reasonable approximation.

Actually, we can prove the following

w

A RELATION

BETWEEN Idm AND

theorem:

has n mutually d i s t i n c t i s given i n x-plane. arg x = 8 matrix p1 , p2 , M , co and an n-bJ-n

Assume t h a t the matrix

11.1:*

Bo(0)

eiRenvalues, and t h a t a d i r e c t i o n : t h e r e exist p o s i t i v e constants T(x,c)

such t h a t

( i ) the components of

- P ~ I . ~ I -X~

(11.11) (ii)

T(x,c)

T(X,E)

a r e holomorphic i n the domain

1 x 1 2 ~ ,

- m ,

IEIIE~

;

admits an a s m p t o t i c expansion W

(11.12)

as

x

T ( x , c ) 5 2 Th(c)x-h h=o tends t o i n f i n i t y i n the s e c t o r

(11.13)

-Plsarg

uniformly f o r (iii)

X - m ,

IcI s c 0 ;

the matrices

T (8) h

are holomorphic f o r

1 E ] L e o , g&

&

To(0)

non-singular; (iv)

the transformation

(11.u)

v = T(x, a ) z

takes system (11.7)

(11.10).

By u t i l i z i n g this theorem we can c o n s t r u c t n

l i n e a r l y independent a r g x =I3

s o l u t i o n s of system (11.7) i n a neighborhood of the d i r e c t i o n :

s o t h a t they are smooth with r e s p e c t t o t h e parameter asymptotic expansions are a l s o smooth i n

c

(11.1) belongs t o this category i f

.

p>,q

.

E

and t h a t t h e i r

The d i f f e r e n t i a l equation

Notice a l s o t h a t Theorem 6.1

i s , i n some way, regarded a s a s p e c i a l case of Theorem 11.1, although, i n Theorem 6.1, some global p r o p e r t i e s of s o l u t i o n s with r e s p e c t t o

x

and

...,

s e v e r a l parameters

(al, am ) a r e s t a t e d a s a d d i t i o n a l r e s u l t s . There i s no general r e s u l t which i s a p p l i c a b l e t o the case when

Bo(0)

has multiple eigenvalues.

an example when

p
.

The d i f f e r e n t i a l equation (11.1) i s such

I n order t o i l l u s t r a t e d i f f i c u l t i e s i n such a

case, l e t us consider (11.1)under the assumption t h a t

*

p
.

If

E#O

,

See Po-Fang Hsieh and Y. S i b u p [ U J . We do n o t present t h e proof of

this theorem i n this book.

The q u a n t i t i e s

e x p l i c i t l y i n terms of eigenvalues o f

p1

and

p,

can be determined

.

For more information, see, Bo(0) f o r example, Y. Sibuya [ 3 9 ; Section 3 , pp 158-1661, Y. Sibuya [ 3 3 ] , Y.

Sibuya [ 3 7 ] and W. Wasow [47; pp 49-87].

See a l s o Section 56 (Chapter 9 ) .

MAIN PROBLEM

45

t h e matrix

has two d i s t i n c t eigenvalues c =0

0 at

value

.

which reduce t o a m u l t i p l e eigen-

Two eigenvalues of t h e matrix B ( 5 , c )

Applying fieorem 11.1 a t

E#

o , we

of (11.3) a r e

can c o n s t r u c t a transformation

v=T(E,c)z which takes (11.3) t o

'i(s , E ) a r e h+(S,o)

where

powers-of

.

5-l

i s a two-by-two diagonal matrix whose diagonal components

T(5,c)

The components of

.

admit asymptotic expansions i n

Set m

If we expand

X*(~,C)

i n powers of

5-l

--1 s i o n s are polynomials i n 1

, the

c o e f f i c i e n t s of such expan-

(except f o r t h e leading c o e f f i c i e n t which

1

fI

is

2 2 f 24, c ) .

B(5,c)

Since t h e components of t h e matrix

such s i n g u l a r i t i e s a t a l l , t h e q u a n t i t i e s

.

Th(c)

do not have

ought t o have s i n g u l a r i -

This means t h a t t h e asymptotic expansion of T(5,r) i n 5 -1 i s n o t smooth a t c = O W e can a l s o show t h a t , i f we t r y t o expand T(5,e) i n powers of s , c o e f f i c i e n t s will have s i n g u l a r i t i e s

ties i n

E

.

powers of

at

s=m

.

I n t n i s chapter we s h a l l consider a problem which f a l l s within

this category. 12.

Main problem.

Consider a boundary-value problem

y " - [Exmtl+P(x)]y=O (12.1)

Y(0) = 1 Y(X)-,O

where (12.2)

c

,

o<x<+=

,

, (x-,+-)

9

i s a real p o s i t i v e parameter, and m-1 P(X) = x m + y x

+ ...+ am-lx+

am

.

4.6

For

bm AND Umtl

A RELATION BE’I”

E

=0

, the

boundary-value problem (12 .l) i s reduced t o y”-P(x)y=O

(12.3)

1

Y(0) = 1

,

o<x
,

y(x)+O

(x+t-)

.

Problem (12.3) has a s o l u t i o n

(12.5)

um(0,a1,

a

*,am)# 0

-

011 the o t h e r hand, note t h a t , i f we change t h e independent v a r i a b l e

x

by

problem ( 1 2 .l) becomes

Since

problem (12.7) has a s o l u t i o n

(12.8)

Y =

Therefore problem (12 .l) has a s o l u t i o n

1 -mt2 --m+l (12 .lo)

Y =

smt3x, c &3,e &3al, mt2 --mtl

--

bmtl(O,c &3,c

If and only i f (12.9) i s s a t i s f i e d .

&?al,

2 --

...,s &3a m ) 2 --

...,e mt3am

The right-hand member of (12.10) has

47

MAIN PROBLEM

an e s s e n t i a l s i n g u l a r p o i n t a t

s=O

.

Nevertheless, we can prove t h e

following theorem:

=RE24

l

a

For piven p o s i t i v e numbers Ro

p o s i t i v e number

co

6,

, there

exists a

such t h a t

if -

(12.12)

1 mS

lim e +-I

1 m+2 --at1 -- 2 uhl( sm+3x,s m+3,c &3a1,. ..,s m+3am) lim Q --m+2 --m t l -- 2 ++O 7

(12.144)

6

U e 1 h M3,, m+3al,

(12.14) are v a l i d uniformly for

...,c M3am)

1

,b

A RELATION BETWEEN

48

0

bMl

26,

lll,(O,al,...,am)l where

AND

i s an a r b i t r a r y compact set i n x-plane.

c o r o l l a r y i s l e f t t o the readers a s an e x e r c i s e .

The proof of t h i s W e s h a l l p r e s e n t the

proof o f Theorem 12.1 i n t h i s chapter.

13.

x =

Asymptotic s o l u t i o n s a t

equation y f f - F ( x , € ) y = o,

(13.1)

.

i-

Let us reduce t h e d i f f e r e n t i a l

P(X,€)

m+l =€x tP(x)

,

to (13.2)

u ' =A(x,E)u

,

By t h e transformation u=T(x,E)G

,

system (13.2) i s taken t o

where

(x,C )

.

i s a s c a l a r function, and

C

1-

h( x ,G ) = zP(x, c)-%

I

In order t o simplify ( 1 3 . 3 ) , set ( 134 )

where matrix.

z = v t g(x,s)& g(x,c)

,

From (13.3) and (13.4) we d e r i v e

i s a two-by-two

constant

ASYMPTOTIC SOLUTIONS

we get

and

c = Then

and

(13.5) where

(13.6)

[;

-3 .

49

A RELATION BETWEEN Qm

50

AND

By u s i n g t h i s f a c t , w e s h a l l prove t h e following lemma.

€d!mLLL

bers.

Let

Ro

Po

9

and

el

be a r b i t r a r y but f i x e d p o s i t i v e num-

Suppose t h a t I

(13 -10)

O < p o q l

.

Then the d i f f e r e n t i a l e q u a t i o n ( 1 3 . 1 ) admits two s o l u t i o n s y2(x, g j

(i) (X,E

the ,al,

yl(x,c)

and

satisf-ying t h e f o l l o w i n g conditions: (j = 1 , 2 )

yj(x,c)

.. , , a m )

a r e holomorphic with r e s p e c t t o

f o r e v e r y complex v a l u e of

x

and f o r

( c,al,

...,am)

t h e domain

(ii)

the

y.(x,r) 3

f o r real v a l u e s o f

( j =1,2) p o s s e s s t h e asymptotic r e p r e s e n t a t i o n s

x

.

I f a r e a l p o s i t i v e number

large, Re[P(x) 3 2 for

l m

7

xo

is sufficiently

&

ASYMPTOTIC SOLUTIONS

Hence

(13.14) and

(13.15) for

xo

(13.16)

< x <+-

o
+ ...+I

Let u s now estimate

I

9

arg

,


'

.

am[( R o

h(x,c)

€1

g(x,s)

and

g'(x,E)

i n domain (13.16).

Note t h a t lF'(x,c)l i n domain (13.16) i f

< (mt1)l

anXm-l

xo i s s u f f i c i e n t l y l a r g e .

IP'(x,€)P(x,4-11 f o r (13J 6 ) .

€lXrn+

Hence

5 [(mel) + Re EI +

4mlx-1

Since

1 we o b t a i n (13.17) and

(13.18)

i n domaln (13.16)

.

Furthermore,

and hence

1

-2-p (13.19)

I g ' ( x , o ) I (L1x

f o r (13.16)' where

5

i s a positive constant.

Consider the matrix

S(X,E) defined by (13.6), and denote by

IlS(x,€)II t h e sum of a b s o l u t e values o f components of

(13.6)' (13.17)' (13.18) and (13.19) imply t h a t 1 -2--m (13- 2 0 )

IlS(X'E)II
2

S(x,e)

.

Then

52

A

i n domain (13.16) i f

xo

RELATION BETWEEN l,jm

b&l

AND

i s s u f f i c i e n t l y l a r g e , where

L2

i s a positive

constant. L e t us now consider system (13.5), and s e t

v=

[

r:]

'

s(x,e) l+g(x,d2

-i

S1&X

,€)

1

s21(X,€

S12(X,€) SZ2(X, €

1

Then system (13.5) i s w r i t t e n as

-1

k-

2 v ' =P(x,c) Vl+ Sll(X,€)V1+

(13.21)

s12(X,€)V2

-1

v ' = - P ( x , € ) 2 v 2 t s21(x,€)v1t s22(x,"v2

9

.

Estimates (13.18) and (13.20) imply t h a t

uniformly f o r (13.11).

f o r (13.16), i f

xo

Furthermore, estimate (13.15) y i e l d s

i s sufficiently large.

If we u s e t h e s e e s t i m a t e s , i t i s n o t d i f f i c u l t t o show t h a t , i f i s s u f f i c i e n t l y l a r g e , t h e r e e x i s t s a two-by-two m a t r i x

(i) T(x,c) (ii) l i m X+Sm

i s well-defined i n domain (13.16); T(x,c) =

1' 1'

l o I!

(iii)t h e t r a n s f o r m a t i o n

Hence

uniformly f o r (13.11);

T(x,G)

xo such t h a t

-

CASE WHEN P(X,€) = E X 2 + x

i s a fundamental matrix of system 1 3 . 2 ) . Lemma 13.1. The Wonskian of t h e two s o l u t i o n s dent

Df

x

.

53

This completes t h e proof of

y1(x,c)

and

y2(x,e)

i s indepen-

Observe t h a t t h e asymptotic formulas (13.12) and (13.13)

yield

Hence

i n domain (13.16). The s o l u t i o n y ( x , ~ ) i s subdominant along t h e p o s i t i v e r e a l a x i s 2 i n x-plane. Notice t h a t t h e asymptotic formulas (13.13) a r e , i n a c e r t a i n

.

and y; i n terms of ? ( x , c ) It 2 i s p o s s i b l e t o f i n d more t e r n s i n these expansions so t h a t w e obtain b e t t e r sense, beginnings of expansions of

r e p r e s e n t a t i o n s of

y2

f i n d the behaviors o f

yz

f i n d a r e the behaviors of

.

y;

and

y

This, however, does not mean t h a t we can

and

y;

as

c

y2

and

y;

as

-P ( x , s )

tends t o zero. (What we w i l l tends t o i n f i n i t y . )

In

t h e following s e c t i o n s , we s h a l l construct a d i f f e r e n t r e p r e s e n t a t i o n of a sub/dominant s o l u t i o n i n such a way t h a t we can f i n d the behavior of the subdominant s o l u t i o n a s

.+!,I

Case when

e

tends t o zero.

2 F(x,g) = e x + x

.

I n 1931, F. Zernike [ 5 0 ] s t u d i e d t h e

d i f f e r e n t i a l equation

(14.1)

y"-

[€X2t x ] y = O

.

He sought t h e s o l u t i o n s a t i s f y i n g t h e conditions

and he showed t h a t , f o r s m a l l the f u n c t i o n

w(x)

0

, such

a s o l u t i o n may be approximated by

satisfying

w ~ ~ - X w = O,

(14.3)

i

w(0) = 1

,

w(x)+O

(x+scD)

.

I n f a c t , he was a b l e t o obtain t h e following r e s u l t :

54

A RELATION

The f u n c t i o n

%: -

BETWEEN Idm

AND Idrntl

y ( x , ~ ) s a t i s f . y i n g (14.1)

and

(14.2) admits

an asymptotic r e p r e s e n t a t i o n

(14 -4)

Y(X,E)

uniformly i n

x

%P(X,€)W(X)+ s q ( x , s ) w ' ( x )

r e a l and p o s i t i v e as

r e a l a x i s , where

p(x,~)g & -

c o e f f i c i e n t s a r e polynomials i n

q(x,E) x

8

t e n d s t o zero along t h e p o s i t i v e

a r e formal power s e r i e s i n

E

whose

.

E v i d e n t l y Z e r n i k e l s problem i s a s p e c i a l c a s e of t h e problem of t h i s chapter.

We s h a l l g e n e r a l i z e t h e method i n his proof of Theorem 14.1

t h a t we may u s e i t i n t h e proof of Theorem 12.1.

Keeping this purpose i n

mind, ue s h a l l e x p l a i n how t o c o n s t r u c t t h e formal power series and

q(x,e)

p(x,s)

i n this section.

Se i,

Y =P(X,E)W(X and der.ive

and

Upon i n s e r t i n g t h e s e r e s u l t s i n t o (l4.1),w e g e t c

e q u a t i o n s ( 1 4 . 6 ) become

x

See F. Zernike [ 501.

SO

-

55

Then (14.8) y i e l d s a system o f equations ”

r

(14.9) and

(n=1,2,

...) .

System (14.9) i s equivalent t o (:o

(14.911

-- q”’(x)+ 2xq;(x) + qo(x) - x2 = 0 , 2p;(x) + q & ( x )= 0 .

Also, system (14.10) is equivalent t o

( n = 1 , 2 , ...)

.

and s u b s t i t u t e (u.11i) n t o t h e f i r s t equation o f (14.9l) t o d e r i v e

- x

This i m p l i e s

r=2

, and

2

=o.

hence 1 2

.

q o ( x ) =-x 5 The second equation of (14.9’) y i e l d s

where

co

i s a constant.

The constant

co

i s determined by t h e

A RELATION BETWEEN Idm AND

56

requirement t h a t

l,u

y(0) =1 or

or

g o , € ) + q(O,r)w'(O) = o

(14.12)

.

Condition (14.12) i s e q u i v a l e n t t o (n=O,1,2

pn(0)+qn(O)wt(O)= O Since

po(0) = c o

, we

qo(0)=O

and

POW

have

co=O

,...) .

, and

hence

1 =-7

Assume . t h a t polynomials

po( x ) ,

...

found by (14.9) and (14.10) f o r

x)

n(h-1

,

qo(x),

. ..,qh-l(x)

have been

t o g e t h e r with ( 1 4 . 1 2 ) . Then

inserting

i n t o t h e f i r s t e q u a t i o n o f (l4.101) f o r

0551.

where

, uniquely.

ch

n=h

, we

find

The second e q u a t i o n o f ( U + . l O I ) ,

r

and

qh,j

,

then, yields

i s a c o n s t a n t which i s determined by C h t

qh(o)wl(o) = 0

.

Thus we can f i n d a formal s o l u t i o n Y =p(x,g)w(x) + cq(x,s)w'(x)

of equation (l4.1) which s a t i s f i e s t h e c o n d i t i o n P(O,E)W(O)+ e q ( o , € ) w ' ( o ) = 1 I n s t e a d of completing t h e proof o f Theorem 14.1, we s h a l l g e n e r a l i z e this method of Zernike i n t h e n e x t s e c t i o n .

15. A g e n e r a l i z a t i o n o f Zernike I s method. the d i f f e r e n t i a l equation

(15.1)

y" - [ axw1

+ P( x) ]y = 0

t o an e q u a t i o n of t h e form (15.2)

w It- ( P ( x ) t ES ( x

by a formal t r a n s f o r m a t i o n

,6) 3 w = 0

I n this s e c t i o n we s h a l l reduce

A GENERALIZATION

(15.3) Acre

y = p(x,e)w+ c q b , e)w'

S

,p

polynomials i n

(x,a)

with r e s w c t t o x

9

w e formal power series in

and q

.

, and

57

E

whose coefficients a r e

furthermore S is a pol.momial of degree

From (15.2) and (15.3) we derive yl =pf(x,c)w+ p(x,e)w'+ eql (x,c)w I t cq( x, B ) w"

= {p'(X,6) -+ Sq(X,G)[P(X)+ ES(X,E)I}W

+ and

equations (15.5) become

[P(X,E)

+ cq'(x,c) Iwl

m-2

A FlELATION BETWEEN ldm AND

58

b ~ l

(15.7)

Set

(15.9)

( n = 1 , 2 , ...)

.

Let us consider an e q u a t i o n

1 -7X"'+ 2P(X)X't P ' ( x ) X t Y

(15 .lo) where

X

and

mials i n

(al,

Y

a r e unknown q u a n t i t i e s , and

...,am )

and

yo#O

m-2 Y(x) =

(15.11)

z

Y,X

.

Set

m-2-h

h=o

If N < m-1

, set X(x)

(15.12) and, i f

(15.13)

NLm-1

N-k = z Ykx k=o

3

0

, set x(x) =

N-Ill+l

z

h=o

3"N-Wl -h

yo,

9

...,yN

a r e given polyno-

A SUBDOMINANT SOLUTION

59

Then (15.lo) i s equivalent t o

(15

N Y(x) = z k=o

.u)

YkX N-k

if

,

N<m-1

and

+ m]Xo = y o , [ 2(N-mtl-k) + "1% = yk t p k [2(N-mtl)

(15 -15)

Yh=YN-mt2+ht% if

, where

N>m-1

(i)

cpk

i s a polynomial i n

(al

i s a polynomial i n

..., N - m t l )

( h = 0 , 1 , **.,m-2)

,...,am,Xo, ...,$-1)

are i n t e g e r s ; ( i i ) Qh

( k = 1,

(a1,

,

9

whose c o e f f i c i e n t s

...,am,Xo, ...,sJ-mtl) whose c o e f f i c i e n t s

are integers. I n this manner w e can determine

qn(x)

and

Sn(x) ,

t i o n of (15.8) and the second equations o f (15.9) determine

, respectively.

pn(x)

The q u a n t i t i e s

pn(x)

.

( n = 0,1,. .)

The second equaand p,(x) contain a r b i -

t r a r y constants which we will f i x by t h e conditions pn(0) = 0

(15.16)

16.

( n = 0,1,.

..) . Let u s write

A subdominant s o l u t i o n of equation (15.2).

S ( X , E ) as

m,U s(x,c)=

z

m-h sh(c)x

h=2 where

(16.1)

.

.

S are polynomials i n (a1,. .,a ) The power series (16.1) h,n m a r e formal. However, by using Theorem 5.1 (Chapter 1), w e can f i n d m-1

and t h e

functions

Ph(a,c)

(i) the

8,

where

, cl

Ro

( i t ) the (16.3)

ph

(h=2,

...,m)

such t h a t

a r e holomorphic i n t h e domain:

and

PO

a r e a r b i t r a r y but f i x e d p o s i t i v e numbers;

s a t i s f y t h e asymptotic conditions

A RELATION

60

uniformly f o r

€1 < P o

law

BETWEEN bm AND

+ ...+lamI L R o

lall

-

as

tends t o zero i n t h e s e c t o r

c

We s h a l l consider t h e d i f f e r e n t i a l equation (15.2) w i t h

(16.A) If we s e t

{

=

(16.5)

(h=l)

+:

,

€Bh(a,d

(h=2,

...,m)

,

we obtain a s o l u t i o n

(16.6)

w = f ( x , a , c ) =bm(x,al(a,c),

of (15.2)

.

.. . , a m ( a , c ) )

This s o l u t i o n i s c e r t a i n l y subdominant along t h e p o s i t i v e real

axis i n x-plane.

Since b m ( x , a ) i s e n t i r e i n (x,al function f ( x , a , g ) a s a power s e r i e s i n

,...,am 1 , we can express t h e (cP,, ...,e~,) . S e t

(16.7)

where

(p1 = p 2 + . . . + p m

and t h e

-.

fm (x,a) ,P2’ * ’Pm

are entire i n

(x,a)

.

From (16.7) we d e r i v e

(16.8) If we s e t

al

(h=l)

,

ch(a,B) = { ah + r eieh where

r

...,m) , t h e oh a r e real v a r i a b l e s ,

(h=2,

i s an a r b i t r a r y p o s i t i v e number and

then

(16.9)

f

mYP2’.

..tPm( x , a )

Therefore, the power s e r i e s (16.7) and (16.8) a r e uniformly convergent for O<x<+=,

(16.lo)

1811 +

*.. bmI9,

O
Y

9

lw!K E I I P o

’

61

PROOF OF THEOREM 12.1 Now i t i s n o t d i f f i c u l t t o show t h a t the s o l u t i o n f tives f 1 admit asymptotic expansions

and i t s deriva-

m

as

tends t o zero i n the s e c t o r

t

(16 -13) where

lug e k P o

fn(x,a)

(16 J 4 ) uniformly f o r

9

are entire i n

i

(x,a)

fn(x,a) I 0

,

f,:(x,a) 5 0

,

I all + ...+IamI
, and ( n = 1 , 2 , ...) as

x

,

tends t o

i-

.*

Note t h a t

of the asymptotic expansions of bh a r e polynom-. the c o e f f i c i e n t s S h,n i d s i n (al,. ..,a m ) , although the ph a r e not n e c e s s a r i l y e n t i r e i n (al,...,a 17.

m)

-

Proof of meorem 12.1.

The r e s u l t s of Sections 15 and 16 y i e l d a

fopmal s o l u t i o n

y = d x , E 1f ( x , a , E 1 + sq(x, E 1f 1 (x,a , c )

(17.1)

o f t h e d i f f e r e n t i a l equation (15.1).

~~

~~

*

Set

~~~

Cf. Theorems 3.1 and 5 . 2 of Chapter 1.

bm

A RELATION BEWEEN

62

AND

Furthermore, m

YN(x,E) ' b m ( x , a ) +

n

C

n=l

(17.6) ?~(X,E)

n

+ c

gU;(x,a)

E yN,,(x,a)

E

9

y;,,(x,a)

n=l

tends t o zero i n s e c t o r (16.13), where t h e

uniformly f o r (16.12) as

E

yN,?(x,a)

(x,a)

are entire i n

'0

( 1 7 .7) uniformly f o r Since 6

'

Y $ , ~ ( x , ~ 0)

9

...t laml L R o

+

lal/

p(x,c)

, and

and

q(x,c)

as

x

tends t o

so0

.

s a t i s f y (15.4) a s formal power series i n

,

(17.8)

yi(x,c)

= where

P N ( x , ~ ) and

holomorphic i n powers of

E

N t2

6

p(x) ]fN(x,

€1

[PN(x,c)f(x,a,c)+ QN(x,c)f'(X,a,E)l

QN(x,g)

(a,€) as

- [Exm t 1

are polynomials i n

x

9

whose c o e f f i c i e n t s a r e

f o r (17.5) and admit asymptotic expansions i n

t e n d s t o zero i n s e c t o r (16.13), uniformly f o r

E

[all t

...+ l a m [< R o .

Set

(17.9)

y=YN(x,C)t

i n t h e d i f f e r e n t i a l e q u a t i o n (15 .l) t o d e r i v e

(17 .lo)

z"

- [ exm+'+

~ ( x1z) = -g

N+2

[p,(x, e ) f ( x , a , c) + %(x,s)fl(x,a,E)l

The d i f f e r e n t i a l e q u a t i o n (17.lO) (17.11)

2

has a s o l u t i o n

= Z,(X,E) N+2

=?

X

[Yz(X,€)

[

X

where

yl(t,€)DN(t,C)dt 0

.

THEOREM 12.1

PROOF OF

and

and

yl

solution

63

y2 a r e the s o l u t i o n s of (15.1) given by Lemma 13.1.

z N ( x , € ) i s holomorphic f o r every complex value o f

i n domain (17.5).

The

x and

(a,€)

Furthermore, w e can prove t h a t

,

Ilm €-N-2zN(x,€) = o (17.13) x+sol l i m €-N-2z$(x,

=o

€)

X++J

uniformly f o r

(a,o)

i n domain (17.5).

I n f a c t , we can prove (17.13)

without any d i f f i c u l t y , i f we use the formulas

-1 -

-4

--1

-1

5,

2 y1(x,e)y2(5,c) =[If o(l)]~(X,e) P ( 5 , o ) 4eXP[-s P ( S , c ) dS1

(17.l.44)

,

X

and y i ( x , e ) y 2 ( 5 , s ) = [1+o ( 1 ) (17.15)

i

-I P ( x , -1c ) ~ ~ --(1S“,x~P)[ -~~5,P ( s , c ) -12d s l -1

y1 ( x, ) y; ( 5 ,

= [-1t ( 1)IF( x,

-1

j4 F ( 5 , )heXp[-

d

,

-1

fP-

( s , ) 2ds I

,

X

together with the estimate (17.16) Formulas (17.l.4) and (17.15) a r e d i r e c t consequences of Lemma 13.1, and x and 5 tend t o $.a (19.16) i s v a l i d f o r s u f f i c i e n t l y l a r g e p o s i t i v e values of x

they a r e v a l i d uniformly f o r (17.5) a s

. .

Estimate

Thus w e constructed a s o l u t i o n (17.17)

y=$N(x96) =fN(x,6)

of t h e d i f f e r e n t i a l equation (15.1).

zN(x96)

This s o l u t i o n i s subdominant along

the p o s i t i v e r e a l a x i s i n the x-plane uniformly f o r (17.5).

Furthermore,

(17.6) and (17.13) imply t h a t

(17.18)

s ~ ( x , E=)b m ( x , a ) + o ( c )

uniformly f o r (16.12) a s

,

*~(x,E)= ~ ; ( x , a ) + O ( E )

tends t o zero i n s e c t o r (16.13).

6

Therefore,

we can complete the proof of Theorem 12.1 by using the uniqueness of the subdominant s o l u t i o n

bmtl

.

Theorem 12.1 i s not a complete generalization of Zernikels r e s u l t (Theorem+!,.I .1)

.

Zernike derived an asymptotic representation (14.4)

.

W e

derived only formulas (12.13) and (12.l.4). However, i n the proof o f Theorem 12.1, we derived a c t u a l l y an asymptotic representation a f t h e solut i o n of the boundary-value problem (12.1).

F.E. Mullin [25] generalized

44

bm AND 4wl

A RECATION BE’I”

Z e r n i k e l s result t o a d i f f e r e n t i a l e q u a t i o n o f t h e form

(17.19)

xPyf’-[ E P ( x ) +Q ( X , B ) ] Y = ~

in t h e c a s e when p = 2 and t h e c o e f f i c i e n t s o f

, where

P(x)

Q(x,g)

and

are holomorphic i n

than two, Z e r n i k e f s method must be modified. of t h e formal power series

p(x,c)

r a t h e r than polynomials i n

x

.

a r e polynomials i n

Q(x,e )

and

E

.

If

p

x

i s greater

For example, t h e c o e f f i c i e n t s

q(x,e)

must be r a t i o n a l f u n c t i o n s

(See F.E. Mullin [25; p 193.)

CHAPTER 4 ASYMPTOTIC EZHAVIOR OF Um(x,a) @ am TENDS TO I N F I N I T F

I n this chapter we s h a l l study t h d d i f f e r e n t l a l equa-

18. Main Droblem. ti on y"

(18.1)

- [ xm + alx m-1 1x1

f o r large value of

+

...+ ammlx + am+A]y=O

, W e s h a l l d e r i v e asymptotic estimates of t h e

function

(182)

Ilm(x+sYalY ..*,'Fin-pm+ A )

for large set i n

.

I A 1 , where

(x,al,

(x,al,. .,am) i s assumed t o be I n a given compact am )-space. The v a r i a b l e s is r e s t r i c t e d t o a r a y i n

...,

the s e c t o r

8,: l a g

<=. n

XI

By v i r t u e of Theorem 7 . 3 (Section 7, Chapter 2 ) , It i s s u f f i c i e n t t o

consider

(18.3)

Um(s,al, ***,am-19am+ A)

i n s t e a d of (18.2).

(18.4)

Let us denote by P(x,A)

m m-1 x +alx

+ ...+am-lx+am+X

t h e polynomial

.

men, t h e d i f f e r e n t i a l equation (18.1) I s equivalent t o t h e system

(18.5)

u'=A(x,X)u

where

*

'Ns chapter I s mainly based on Y. Sibuya [ 3 4 ] .

66

ASYMPTOTIC B M A V I O R OF Um(x,a)

As i n Section 1 3 (Chapter 3 ) , the transformation

1

(18.6)

reduces sys t e m ( 18.5 ) t o

(18.7)

v ' =B(x,b)v

with

(18.8) where

19.

Estimates i n t h e s e c t o r

larg A1

5 TT-

p,

.

In this s e c t i o n we s h a l l

consider t h e d i f f e r e n t i a l equation (18.1) i n the domain

(19 .li

Ro

where that

p,

and

1

larg A (

In- Po

9

o(_x<scD,

tall

Po

+

. . a +

laml 'Ro

f

a r e a r b i t r a r y but f i x e d p o s i t i v e c o n s t a n t s .

is s u f f i c i e n t l y small. ro and

There e x i s t two p o s i t i v e numbers po

and

Ro such t h a t lP(X,A)l

(19.2)

W e assume

Ip(x,a)l

2rolxI m

2r0x

l=-g[P(x,h)ll

9

,

i n - 21

Po

Mo

which depend only on

Iarg L I

ESTIMATES I N

TI

- p,

67

f o r (19.1),i f 1 h 1 2Mo

(19.3)

*

In f a c t , t h e r e exists a p o s i t i v e number

m

Jarg[x +alx

for x l x ,

,

(all

+ ...+I

m-1

xo

such t h a t

+ ...+am]l < - p1 2 0

aml L R o

.

Therefore, i f

X ~ X ,

, we

have

I arg[P(x,h) 31 In - Po for (19.1).

(See Flg. 19.1.)

Furthermore, i f

Mo

i s sufficiently large,

we have

1 arg[P(x,X) 3 I L f o r (19.1) i f

2Mo and

1

17

- ?Po

O<x(xo

.

(See Fig. 19.2.)

-%Pe

CFT

c-T --

A - plane fig. 19.1.

68

ASYMPTOTIC BEHAVIOR OF um(x,a)

satisfies (19.3). Then, t h e r e exists a p o s i t i v e constant l m 1 m
A

.

1112Mo

9

l a g AII"-Po

9

and o<x

m

15,IXl

9

t h e r e exists a p o s i t i v e constant

a a

2-

IP(X,UI

such t h a t m x

9

50

since p ( x , a ) =Ixl[$mtp -1alx-m-1

+ ...+w-ma

,

+eie]

m

Thus (19.2) holds i n domain (19.1) i f

h

s a t i s f i e s (19.3) and

Mo i s

sufficiently large.

Assume (19.1) and (19.3).

get

Then, i f

Mo i s s u f f i c i e n t l y l a r g e , we

-2-Zp 1

where

S(x,X)

i s matrix (18.9);

x)_l

for

O<x<xo

for

x2l

for

O(_x<xo

,

, ,

IlS(x,X)(l i s t h e sum of absolute values

S ( x , l ) ; g(x,X)

of components of

,

for

i s t h e function which i s given by the

K i s a p o s i t i v e constant which depends only and Ro ; xo i s a real p o s i t i v e v a r i a b l e ; K(xo) i s a p o s i t i v e

second equation of (18.10);

on

p,

constant which depends only on xo independent of

X

,

we can d e r i v e

, p,

and

Ro

.

Since

is

(19.4) from (19.2) without any d i f f i -

culty. Estimates (19.4)

P'(x,h)

imply t h a t t h e two q u a n t i t i e s

69

as

1 tends t o i n f i n i t y i n the secto r

(19.8)

l a g A1 I n - P o

9

Consider system (18.7), and a e t 1 X (19-9)

v=w

-2

exp[-l P(T,X) d ~ .] 0

Then 1

70

ASYMPTOTIC BEHAVIOR OF Um(x,a)

If

-19.1:

Mo> 0

i s s u f f i c i e n t l y l a w e , system (19.12) has a solu-

tion -

i n t h e domain defined bx (19.1)

and

( i ) C1(x,A)

(19.15)

as

C (x,h) 2

lal[ t

w ( x , h ) = 1 t G2(x,h)

(19.3) such t h a t

tend t o zero uniformly f o r

...t l a m [I R o

,

O<x
tends t o i n f i n i t y i n t h e s e c t o r

(19.8)

l u g hl 9 - P 0 ; G2(x,X)

(ii) ,Sl(x,h)

as -

, and

y ( x , X ) =G1(x,h)

(19-14)

tends t o

x

Sa,

tend t o zero uniformly f o r

.

This lemma implies t h a t system (18.7) has a s o l u t i o n v(x,X)

=[

Gl(X,A)

]

-1

X

exp[-J

1-tG2(x,X)

.

P(T,A)~~T] 0

S u b s t i t u t i n g this s o l u t i o n of (18.7) i n t o (18.6), we o b t a i n a s o l u t i o n of equation (18.1) of t h e form

--1

I

(19.17)

where

Fl(x,X)

( i ) Fl(x,h)

-1

X

y(x,X) = [ l t Fl(x,A) ]P(x,A) 4 e ~ [ - Jp ( ~ , A ) ~ d ,~l

-

l o X 1 y ’ ( x , X ) = [-1+ F2(x,A)1P(x,h)4exp[-J P ( T , A ) ~ ~ T,I 0

and

and

F2(x,h) F2(x,X)

s a t i s f y t h e following conditions: tend t o zero uniformly f o r (19.15) a s

A

tends

t o i n f i n i t y i n s e c t o r (19.8); (ii) Fl(x,A)

tends t o

+

Since

and

.

F2(x,h)

tend t o zero uniformly f o r (19.16) as

-1

2 Re[P(.r,h) ] > 0

positive real axis.

Hence

bm(x,al

, solution y(x,A)

,...,am-1’

(19.17) i s subdominant along t h e

i s a constant m u l t i p l e of am t A )

.

We s h a l l determine this constant m u l t i p l i e r . ( 19.18)

x

bm(x,al

To do this, set

,...,am-1’ am +A)=tm(al ,...,am-1’

a +A)y(x,h)

m

.

ESTIMATES IN

As

x

tends t o

+a0

, we

Im g

5 n-

71

po

have

i -&

where

A(

(m: odd)

,

and -h h=l The q u a n t i t i e s

bh

and

Section 6 , Chapter 2.)

If

r

m

a r e polynomials i n

al

,...,am-1’

a +?,

m

m i s odd,

Theref ore (19.20)

t m ( al , . . . , a m-1’ am + A )

h=l

(m: odd) If

m is even,

Theref ore,

.

.

(Cf.

ASYMPTOTIC BEHAVIOR OF bm(x,a)

72

c)

( 19.21

Cm(al,

...,am-1'

am + h )

(m: even) Set

-

f i P ( ~ , h )2 -

T

h(al, ".,a

1- h 7

- $ n tCt l )

\T

] d ~

h=l

0

(19.22)

7

.

(m: odd)

,

m+A) = h=l

0

(m: even) m=l

In p a r t i c u l a r , i f

.

,

J(m+ 2) = 3/2 2

and hence (19.23)

%(al+A)=J

--1

1.1

$Q

[ P ( T , A ) 2 - ' f 2 - L (2 ~

1+ A ) T

0

If

(19-24)

la1 2Mo

9

larg X I 9 - P o

9

and Mo i s s u f f i c i e n t l y l a r g e , t h e r e are no singular poin i n the s e c t o r

9

of

P(

0 5 a r g T<arg A or 02-g

T2-g

A

.

Therefore, by v i r t u e of t h e behavior of t h e integrand of (19.23) a t and

, we

can change T=ht

t o obtain

T

by (O(_t<+)

T=O

ESTIMATES IN

=

2 ,

73

( a r g A 1 I n - po

[ ( t + 1+ A-’al)

1 1 - t2- $( 1+ A-’al)

--1 t

]d t

.

0

Observe t h a t

uniformly f o r

(al,A)

t

as

tends t o

-1 1- . (t+$t

as

tends t o

.

i-n

, and

+cD

that

-2.

t 2 = O ( t 2,

Hence

1 1

2,

1

% ( a l + h ) = h 2 [ J ( ( t + 1 ) 2 2- -5t t1

-5)dt+O(h-’)]

0

uniformly f o r

al

as

X

tends t o i n f i n i t y i n s e c t o r (19.24).

1 1

-t=

[(t+1)2-t

2

1

-2-5 ]at=--2 1

3

0

Notice t h a t

.

Thus

2 \

uniformly f o r

al

If

, we

m=2

as

h

tends t o i n f i n i t y i n s e c t o r (19.24).

have

5(m+2) I =2 ,

,

P(T,A)=(T 2 +a17+a2)+h 1

and hence i-

(19.26)

-1

L2(al,a2+ A ) =J [ p ( T , h 1 2

-

0

If

A

i s i n s e c t o r (19.24), and

*

.

Mo i s s u f f i c i e n t l y l a r g e , t h e r e a r e no

-1 s i n g u l a r p o i n t s of

1 1 2 -(a + A ) --a 2 2 7 - p1T+1 1

P ( T , ~ )i ~ n the s e c t o r

ASYMPTOTIC BEHAVIOR OF Um(x,a)

74

or

.

1

O>_arg 7 2 , a r g A Therefore, we can change -1 2 T=A t

T

by (O
t o obtbin

1 1 2 1 1 1 1 -(a + A ) --a + O l 2 2 2 2 1 2 2 = J [ ( k t t x a1 t+a 2+ A ) - A t - -2a1 ']k2dt

-1-

0

'

xLtt 1

+

2

= Xs [ ( t tl+alX

-1

-1

2

-1 -1 2 t + A a2) - t

- -A12 2al

0

1t k-4

-

$a2

- Fl) 1 2

2

t+

x

--1 2

]dt

Observe t h a t

1 -1 2 -2 -1 2 [ ( t t l + A alt+h a2) - t

+co

J

1,A-j-a 1 1 2 2 2 2-25) Idt t + l

-a:1-2a1 -

0

= o(1) uniformly f o r (19.27) as

X

la11

+

la21

Go

tends t o i n f i n i t y i n s e c t o r (19.24).

uniformly f o r (19.27) a s

I n the same way, f o r

A

Hence,

tends t o i n f i n i t y i n s e c t o r (19.24)

m 2 3 , we

.*

get

u -+a) = h 2 m[~p)+"11 1

(19.29) uniformly f o r

Lm(al,

...,am-1' am

I all + .. .+ I a,1 5 Ro

(19.24), where

*

See also Section 23 (Chapter 5 ) .

O(X

as

h

tends t o i n f i n i t y i n sector

.

75

1

1

?P [ ( t m + 1 ) 2 -]td t > O .

SCD Lm (O)=S 0

Thus, we have proved t h e following theorem.

mORW 19.1: Assume t h a t

where Ro -

I

1 In- Po larg 1

9

O<X<+,

1J.

+

lam[I R o

. . a +

9

a r e a r b i t r a r y but fixed p o s i t i v e constants. t h e r e e x i s t s a p o s i t i v e constant Mo such t h a t , if A1 2Mo , po

I

um(x,al,...,a

m

(i) P ( x , ~ = ) x

+ alx m-1 151+

as

X

(1.)

...+ am-lx+

am+ 1 ;

tend t o zero uniformly f o r

...+lam[<

F2(x,A) laJ+

x

am + A )

R ~,

tends t o i n f i n i t y i n the s e c t o r

(iii) Fl(x,h)

as -

+

F2 ( x , h )

(ii) Fl(x,A)

m-1’am + A )

...,am-1’

b;(x,al,

where -

Then,

tends t o

-@

a * . +

05x<+a,

1 arg 1I 5 n -

;

tend t o zero uniformly f o r IamI < R o

9

lwg

A I In- Po

;

2

‘ii1( a1+ a )

p,

-1 2 2 =exp[-TA ( ~ + o ( A ) ) I

,

7

14 2Mo

76

ASYMPTOTIC BEHAVIOR OF l,j,(x,a)

uniformly f o r ( a r g hl ( n - p,

20.

Iall t ... t I aml < R o as

X

tends t o i n f i n i t y i n t h e s e c t o r

, where

Estimates i n t h e s e c t o r

l a r g 1-nl (p,

.

Consider the d i f f e r e n t i a l

equation (18.1) f o r

where

Ro

and

a r e a r b i t r a r y but f i x e d p o s i t i v e c o n s t a n t s .

PO

Let us

assume t h a t n

O < P 0 < X

(20.2)

W e s h a l l again denote by

*

P(x,k) polynomial (18.4).

?'here e x i s t two p o s i t i v e numbers

Ro

, such

ro and

Mo

, which

depend only on

that

for (20.1) i f

bl 2 M o

(20.4)

-

' h i s can be proved i n t h e same way as w e proved t h e f i r s t and second ine-

q u a l i t i e s of (19.2). men, i f

(20.5)

where

Assume t h a t (20.1), (20.2) and (20.4) a r e s a t i s f i e d .

Mo is s u f f i c i e n t l y l a r g e , we g e t

I

S(x,A)

1

-2-7l

for

521 ,

IIS(X,A )I1 5 K(So)

for

OlS15,

1 g ( x , h ) I 2%

for

521 ,

for

OL555,

IIS(X,X )I1 L K5

IxI -1-F 1

1 g(x,h )I A m o ) I A 1 i s matrix (18.9);

of components of

S(x,X) ; g(x,X)

-2 9

2

-2

IlS(x,l)ll

9

is the sum of a b s o l u t e values

i s t h e function given by t h e second

ESTIMATES I N equation of (18.10);

5,

5 po

l a r g A-nl

77

i s a p o s i t i v e constant which depends only on

K

i s a r e a l p o s i t i v e variable;

5, and Ro way as we derived (19.4). depends o n l y on

.

Ro ;

K(FO) i s a p o s i t i v e constant which Estimates (20.5) can be derived i n the same

Consider system (19.11), and l e t us transform (19.11) t o a system of i n t e g r a l equations which i s s i m i l a r t o (19.12).

However, i n t h e p r e s e n t

case, t h e path o f i n t e g r a t i o n i s taken along the l i n e

I'*>

(20.6)

(055<+-J) *

x = 5 exP[i

We s h a l l f i r s t d e r i v e an e s t i m a t e of

ii

arg[P(x,X) d x ] along l i n e (20.6).

Set P(x,X) = x m [ l + Q ( x ) ] + and n = a r g [ l + Q(x)] There e x i s t s a p o s i t i v e number

. xo

, which

depends only on

for (20.7) Theref o r e ,

f o r (20.7)

if x

i s on l i n e (20.6).

Hence, f o r such

x

, we

get

Since + n I xn t a r g

n<--n

m+2 we have

Po

A<&+=<$,

Ro

, such

that

ASYMPTOTIc BEHAVIOR OF

78

n

sin(-$ arg A ) m+2 where

bo

urn(

X,

a)

< -6, ,

i s a p o s i t i v e c o n s t a n t depending only on

p,

.

Therefore,

N X ) = o ( l a l 1/m) as

X

tends t o i n f i n i t y i n the s e c t o r

(20.9)

IT-

po<arg ? , < n

.

I t i s easy t o s e e t h a t , i f

Mo i s s u f f i c i e n t l y l a r g e , Im:P(x,X)exp(i&) ] > o

(LO.lL?)

for n

x = 5 exp[i- 2(mt2)1

(20.11)

i

r - p0 Larg h l n la1\ t

... + l a m ]I

52Tl(h)

?

,

lXl,Mo

R ~

9

,

.

Let us Yonsider a system o f i n t e g r a l equations

x

m

where

(x,?,,al,

.. .,a,)

i s i n domain (20.11), and t h e p a t h of i n t e g r a t i o n

i s taken along l i n e ( 2 0 . 6 ) . X

Re[J

(20.13)

1

r

From (20.10) we d e r i v e

-1 P ( ~ , ? ) ~ d
t f o r (20.11), i f

0 = 5 exp[i- 2(;)1 Hence, if Mo

(1x1 L

Slltl)

-

i s s u f f i c i e n t l y l a r g e , system (20.12) has a s o l u t i o n

w,(x,X)

(20.14)

,

=Gl(x,X)

w,(x,X) = 1 t G2(x,h)

f o r (20.11) such t h a t ( i ) G1(x,X)

and

(20.15) as

5

G2(x,b)

tend t o zero uniformly f o r

n- ~ ~ 5 a 1X - g< n

tends t o

i-

,

)A[

LM, ,

)all + . . - + laml I R ~

;

( i i ) t h e r e a r e two p o s i t i v e numbers

L

and

u

such t h a t .

1 a r g A-nl 5 po

ESTIMATES IN

(20.16)

,

IGl(x,h)l I L I A ( - '

79

IG2(x,X)I <_LIAl-D

f o r (20.11). Consider system (19 .ll) f o r n

1

(20.17)

x = s exp[i- Z(lnt2)I n - p0sarg

air

la1( -t ...+

Ia,I

'

0<_55'V(A)

l ~ u0 l , '-Ro

9

t o estimate s o l u t i o n (20.U) for (20.17). Changing

-1 x = J x )m7 exp[i&l

(20.18)

3 sin(&targ

w5{-

X)

s i n (n) 4 1

we d e r i v e

by

9

where

( 2 0 -19)

x

l1 E

'

-12

-1 dwl/d.r = Ih~exp[i&][ZP(x,h)

wl+ sll(x,X)wl-t s12(x,~)w21,

-1

(20.20)

dwJdT= ( a ( m e x p [ i ~ l [ ~ 2 1 ( ~ , h ) w s22(x,X)w21 l+ from (19.11). respect t o

A

Since (20.19) i s an i n t e r v a l which i s bounded uniformly with

, it

(20.21)

Mo

follows t h a t

11

IGl(x,A)(

here

f o r (20.17),

,

+

} l + G 2 ( x , h ) l I L 1 exp[c(hl

?GI

L1 and c are c e r t a i n p o s i t i v e numbers.

Thus, i f

i s s u f f i c i e n t l y l a r g e , we obtain a s o l u t i o n

--1 y(x,X) =[I+ Fl(x,X)IP(x,X) 4exp[-J

-l

{

(202 2 )

-12

X

P(t,X) dt1

oX

1

2 y'(x,X) = [-1+ F2(x,X) ] P ( ~ , h ) ~ e x p [ - S P ( t , h ) dt! 0

of the d i f f e r e n t i a l equation (18.1) such t h a t

11

(i) L2

+ l ~ ~ ( x , h< ) lL

IF~(X,A)~

and

c

3-

exp[cla12 ~ m~

f o r (20.1) and (20.4), where

a r e c e r t a i n p o s i t i v e numbers;

(ii) Fl(x,A)

and

F2(x,A) tend t o z e r o uniformly f o r

lall

+

,

...+lam[ I R ~

ASYMPTOTIC BEHAVIOR OF l,jm(x,a)

80

as

5

,

tends t o

i-= if

i s f i x e d i n t h e domain

A

n-po<arg A l n

, l A l >-Mo .

By using the same method a s i n Section 1 9 , we can prove t h e following theorem.

mom 20.1:

Assume t h a t

1

( 2 02 3 )

where Ro

n-polarg A l n

,

I*>

x = 5 exp[i

blI+

* a * +

055<+-

9

I,.\

IRo

9

a r e a r b i t r a r y but f i x e d p o s i t i v e numbers.

p,

Suppose t h a t

n O < P O < E

E,t h e r e

e x i s t s a p o s i t i v e number

n-

tm a r e

(v)

m23

~ ~ 5 -A g I ~,

Mo such t h a t , if 1x1 2Mo

,

la1 2Mo ;

similar t o those i n Theorem 19.1,

and i n p a r t i c u l a r ,

f

11

-3-

m1

for

ESTIMATES IN

Iarg

A-nl

5 po

81

for -

lal[ +

where

g&

...+

Iaml L R o

,

TI-

po<arg

,

X ~ T I

E are c e r t a i n p o s i t i v e numbers.

We can derive a similar r e s u l t f o r

- n < a r g A(-.+

po

.

From Theorems 7.3, 19.1 and 20.1, we can conclude t h a t t h e function bm(x,al,...,a i s an e n t i r e f h c t i o n o f order

m-1’ am + I )

$+:

with r e s p e c t t o

h ,

CHAPTER 5 STOKES MULTIPLIERS

21.

Preliminary r e s u l t s .

I n t h i s c h a p t e r , ve s h a l l i n v e s t i g a t e connection

Id

formulas f o r t h e subdominant s o l u t i o n s

m,k

(x,a)

.

As a p r e p a r a t i o n , we

s h a l l make a few remarks.

LBM 21.1:

The f u n c t i o n

m,k

(x,a)

admits an as.ymptotic r e p r e s e n t a t i o n

(21.1)

x tends t o i n f i n i t y i n

uniformly on each compact set i n a-space as closed s u b s e c t o r of t h e open s e c t o r s

any

%y%v%+l , where (m:

odd) ,

(m:

even)

(21.2) k+l -m+ (-1) bl

.

T+l

Proof:

Lemma 21.1 follows from Theorem 7.1 i n t h e following way.

Let us

set

rm= rm( a)

bh = tj(a )

and

t o i n d i c a t e t h e dependence of t h e s e q u a n t i t i e s on

.

(al,. .,a m )

l e t us s e t

(21.3)

G(a) = ( w

-1a l , w -2 a 2 , .

. .,w -m am)

and (21.4)

k

G (a) = ( w

-k

al,w

-2k

a2,

...,w -mk am) .

Then, i t i s n o t d i f f i c u l t t o show t h a t

(21.5) Therefore

bh(Gk(a)) =w-&bh(a)

(h=1,2y...)

.

.

Also,

PRELIMINARY RESULTS

,

(m: odd) (21.6)

1-5-

(-1)k bl

,

(m: even)

=pl(a) and

k k x,G ( a ) ) ~ ( - 1 Em(x,a) )

-k

Em(”

(21.7)

.

I t i s e a s y t o d e r i v e (21.1) and (21.2) from ( 7 . 4 ) , (21.6) and ( 2 1 . 7 ) .

In

p a r ti c u l ar ,

.

k r =rm(G (a)) m,k

(21.8)

The two s o l u t i o n s bm,ktl(x,a) Therefore,

Set

*

bm,k (X

(21.9)

) = $( a

9

)urn,k + l (

i s a connection formula f o r -This % ( a ) a r e t h e Stokes m u l t i p l i e r s multipliers

Ck(a)

THEOREM 21.1:

-Ck(a)

and

Set c(a) =Co(a) ,

( 2 1 .lo)

Then, c(a)

and

, for

9

a)

’-‘k ( a )urn,k+2(

9

a)

and t h e c o e f f i c i e n t s

um,k

with r e s p e c t t o

(iii) t h e f u n c t i o n

a s functions of

and

*

Ck(a)

and and

,

‘(a) =

...,am .

...,am ’ -Ck(a) = ? ( G k (al, a)) ;

?(a)

-

al,

-c(a) =C0(a) .

a r e entire i n

?(a)

(ii) $ ( a ) = C(Gk(a))

(21.11)

\dm,ktl ( x , a )

We s h a l l s t u d y , i n this c h a p t e r , v a r i o u s p r o p e r t i e s of t h e Stokes

b m ,k+2

(i)

a r e l i n e a r l y independent.

i s a l i n e a r combination o f

bm,,(x,a)

bm,k+2(X,a)

and um,kt2(x,a)

*

has t h e following form:

{::

1-2w(a)

(m:

odd) ,

(m:

even) ,

where (21.12)

Proof:

w(a) = b l

ptl

(a)

.

Let us s e t bm,k (

(21-13)

9

a)

um,h(

9

a)

Wk,h(a) bA,k(x, a )

u;,

h(x, a )

It i s e a s y t o v e r i f y t h a t t h e right-hand member of (21.13) i s independent

84

STOKES MULTIPLIERS

of

x

.

Since

it follows t h a t

(21.15)

(21.16) Since

and a r e l i n e a r l y independent, t h e Wonskian W1,2(a) m,l does n o t v a n i s h f o r any a Therefore, C(a) and t ( a ) are e n t i r e i n Q

.

a

.

‘This proves ( i ) . To prove ( i i ) ,n o t e t h a t

( 2 12 7 )

Um,k(x,a) =Irm(w

-k

k

x,G ( a ) )

.

(Cf. ( ’ 7 . 3 ) . )

Then, i t f o l l o w s from (21.14) t h a t ( 2 1 .la)

k k b m , k ( x , a ) = C ( C (a) )Um,k+l(x,a) + “ C G (a))Ym,k+2(x9a)

*

“ h e r e f o r e , ( i i )of Theorem 21.1 f o l l o w s from ( 2 1 . 9 ) and ( 2 1 . 1 8 ) .

To prove ( i i i ) ,observe t h a t

r

--

(21.20)

W

0,1

(a) =

1 Xp

r

--

m,lx m++

O(x 2 ) ]

-r

x m , o [ol(+ x 2)]

--1 r m q - l + O ( x 2)]

W

1 --1 -r -mtr w m,l X mJ[l+O(x

53

9

PmLIMINARY RESULTS

85

u l , . . .,u a r e polynomials i n ( x , a ) , and indepenm . ( C f . Theorem 7.3 .) Let us denote ul,. .., um by u1 ( x ,a) ,. ..,urn( x ,a) . ~e t (21.24) u ( x , a ) = (u,(x,a) ,...,um(x,a)) . Proof: -

The q u a n t i t i e s

dent of

s

Then, i t follows from (21.22) t h a t

\( w-kx,Gk( a ) ) = w-&\(

( 2 1-25)

x, a )

...,m) .

(h=l,

Hence (21.26)

Gk ( u ( x , a ) ) =u(w-kx,Gk(a))

.

Observe that (21.l4) y l e l d s (21.27)

b m ( x t s,a) = C(a)bm(w-1 (x+s) ,G(a)) t E(a)bm(u-2 ( x t s ) ,G2(a)

.

Therefore, it follows from Theorem 7.3 t h a t (21.28)

~,(x,a)y,(s,ui

= c m m ( u - l x , G ( a j )y,(u

+ where

Km(x,a) i s given by (7.10).

-1s,u(w -1x,G(a

1)

-2 2 -2 "Ca)Km(w x,G (a))bm(w ~ , u ( w - ~ x , G ~ ( a,) ) )

By v i r t u e of (21.26), we can write

(21.28) as ( 2 12-91

Therefore, (21.23) follows from (21.291,

(7.10) and (21.7).

the proof of Theorem 21.2.

lIiWU2Ld

If we set

then (21.31)

where l2 is

s,,(a)Sm(a).

..sl(a)so(a) = l2 ,

the two-by-two i d e n t i t y matrix.

This completes

STOKES MULTIPLIERS

86

Proof':

The connection formulas (21.9) can be w r i t t e n a s

'%,k ( x , a ) ,kjrn,k+l(x,a)1

(21.32;

[bm,k+l( X a ) ,Urn, k+2 ( x , a ) lsk (a

. . ., m t l ) .

( k = 0,1, Xcte t h a t

b m (x,a)

i s single-valued i n

(x,a) -space.

= bm,o ( x , d ,

(21.33:

Hence

-

Um,m+3(x,a)= b m , , ( x , a )

'Theref:re, h

,

=

O

( x , a ) ,bm,l(x,a)

1

.

[Um , 0 i x , a ) ,

~ ~ , ~ ( xlsmt,(a)srn(a). . a ) .sl(a)so(a)

.

This proves 'Theoren; 21.3. THEOREM 21.4:

For each f i x e d

(21.34)

Proof:

, there

e x i s t s an i n t e g e r

k

such t h a t

.

i$(a,#O

Suppose for a c o n t r a d i c t i o n t h a t

ck(a ) = 0

( 2 1 *15:' f o r some fixed

a

.

...,mtl)

(k=0,1,

bm,k

Then, i t follows from (21.9) t h a t

a r e l i n e a r l y dependent. (i) m

a

and

Hence,

must be even, ( h = 0,

(ii)

. . . ,?)1

a r e l i n e a r l y dependent,

Ulcl

.. .,?)1

iiii) brn,zh+l

( h = 0,

a r e l i n e a r l y dependent.

By u t i l i z i n g t h e asymptotic r e p r e s e n t a t i o n s o f i n f i n i t y , we can conclude t h a t

kj m,o

bm,l

and

bm,k

as

x

tends t o

admit r e s p e c t i v e l y t h e

asymptotic r e p r e s e n t a t i o n s

--2 7

J.

r

( x , a ) = x m ' o [ lO(x t

as

x

tends t o i n f i n i t y i n any d i r e c t i o n i n x-plane.

Since m

(6.9) .)

)]exp{-Em(x,a)]

i s even, t h e f u n c t i o n

,

( C f . Lemma 21.1.)

Em(x,a) i s a polynomial i n

x

.

(Cf.

Therefore, two f u n c t i o n s

,

bm,o ( x ,a ) exp {Ern(x a ) } are e n t i r e i n

x

.

and

b,, l(x, a ) exp {-Em(x, a

3

m e asymptotic r e p r e s e n t a t i o n s (21.36) imply t h a t t h e s e

CASE m = l

AND

87

CASE m = 2

two e n t i r e functions must be polynomials i n

, and

that

r

and m,o However, this i s impossible, s i n c e

must be non-negative i n t e g e r s .

x

r m,l

1 'm,o+ 'm,l=-?

*

This completes t h e proof of Theorem 21.4.

m=l

22. al

.

and case

.

m=2

m=l

If

, there

i s o n l y one parameter

If we s p e c i a l i z e Theorem 21.2 t o this case, we have

.

C(al+ x) = c(a,)

(22.1) Therefore,

i s independent of

C(a,)

21.1 t h a t S ( a 1 ) =c(al)

(22.2) where

-Ck(al)

.

Then i t follows from Theorem

(k=0,1,2)

=-w

9

.

w = e x p { T2 n i l

(22.3)

9

al

I d e n t i t y (21.31) becomes

Then, we g e t c(al) = l + w

(22.5)

. m a :In (22.6)

case

m=l

2 ( = -w )

,

C,(al)=l+w

and -

-$(al)

(22.7)

.

( = -w 2 )

(k=0,1,2)

=-w

(k=0,1,2)

.

In particular, (22.8)

This i s a well-known formula f o r Airy's i n t e g r a l

Ai(x)

.

M. Abramowitz and L.A. S t e p [l; p . 446, 10.4.7].) If

m=2

w0

are two parameters

w=exp{-

(22.9)

If

, there

2

4

ni}=i

set

(22 .lo)

1 1 2 b = - a --a 2 2 81'

.

a1 and

a2

, and

( C f . ( 8 . 5 ) , and

STOKES MULTIPLIERS

88

then (22.11)

bl

(a) = b

.

P+l Therefore,

-C(a1'a 2 ) = - u ' - ~ ~ =(-i)exp{-bni] .

(22.12)

( C f , meorem 21 .1,) If we s p e c i a l i z e

1 2 1 E ( x , a ) =-x + ax m 2 2 1

(22.13)

, we

(6.6),(6.9) and (21.22) t o t h e case m = 2

get

,

and

(22.14) If we choose

x

u1=2x+al,

2 u = x + a 1x + a2 ' 2

so that

, then

u2 = 2 b

ul=O

,

>

I

-

L

E ( x , a ) =--a m 8 1

*

Therefore, i t follows from Theorem 21.2 t h a t 1 z

.

C(a a ) = C ( 0 , 2 b ) e x p { a } 1' 2 4 1 Furthermore, Theorem 21.1( ( i i ) )implies t h a t (22.16j

1

C(0,2b) = Co(0,2b) = C2(0,2b)

1

(22.17)

9

C(O,-2b) = C1(0,2b) = C (0,2b

3

Then, by u t i l i z i n g Theorem 21.3, we o b t a i n C (0,Zb) C( 0 , -2b) = ( 21) COS (bn

(22.18)

I n f a c t , i d e n t i t y (21.31) becomes A=A-~

,

where

This means t h a t C(0,2b)C(O,-2b) =

(22 .lSl)

2ni

r(5-tb ) r (12 - b)

The asymptotic r e p r e s e n t a t i o n s of

8-y To"&,, and Slug2"

8-1 ,

m,o respectively.

s e c t o r s cover x-plane completely. and i s v a l i d as

l~

and bm,2 Since m = 2

(Cf. Fig. 8.4.)

are valid i n

, these

two

Therefore, i f

are l i n e a r l y dependent, t h e asymptotic r e p r e s e n t a t i o n of x

tends t o i n f i n i t y i n any d i r e c t i o n i n x-plane.

IJ

m, 0 Hence, t h e

CASE m = l

AND CASE m = 2

89 x , and

e n t i r e function

(x,a)exp{E ( x , a ) ) becomes a polynomial i n m m,o becomes a non-negative i n t e g e r . Note t h a t

r m,o

(22.19)

This means t h a t , i f Similarly, i f

C(0,2b) = O

-(z-b) 1

, then

C(O,-2b) = 0

Therefore, (22.18’)

1 , then -(2+ b)

i s a non-negative i n t e g e r .

implies t h a t , i f we set

1 --ni

&

the function of

b

.

f(b)

, and

i s entire i n b

f(b)

i s one

L+L1 2 m

Therefore, the f’unction

,

does not vanish f o r any values

Note t h a t

. log f ( b ) must be l i n e a r i n

f ( b ) = e x p ( a b + p)

(22.201) a

C(0,2b)r($+b)

Furthermore, the remark given a t the end of Chapter 4 implies t h a t

t h e order of

where

e

f ( b ) =-

(22.20)

i s a non-negative i n t e g e r .

and

p a r e constants. 1

(22.21)

a = l o g 2 - -2 n i ,

b

, i.e.

, In the next s e c t i o n , we s h a l l show t h a t

p=o

*

I f we accept this r e s u l t here, we have proved the following theorem.

THEOREM 22.2:

In case m = 2

, 1 1

(22.22)

where (22 .lo)

i

bFle C0 ( a1 ’a2 ) = C 2 ( a1 , a 2) = 2 e

-in(-b--) 2 4

2n

r($b)

-k2i n ( L2 s t4l )

Cl(al,a2) = C3(a1,a2) = 2-be 4 l e

.. Co(al,a2)

’

J-

2n 1 r(2-b)

9

-bni

= E 2 ( a1’a2) = ( - i ) e

-Cl(al,a2) = C.. ( a

bni 3 1’a 2) = ( - i ) e

9

,

1 1 2 b=p2-rl.

Formulas (22.22) can be a l s o derived from the well-known connection formul a s of parabolic cylinder functions.

( C f . (8.11), and M . Abramowitz and

I . A . S t e p [l; p . 687, 19.4.6 and 19.4.71.)

90

STOKES MULTIPLIERS

23.

Properties a t

a=O

, we

.* If we

a=O

obtain

$(O) =C(O)

(23.11

,

s p e c i a l i z e ( i i ) o f Theorem 21.1 f o r

-% ( O ) = t ( O )

,

(k=O

,...,m t l ) .

'Then, (iii) of Theorem 21.1 i m p l i e s t h a t

-

(23.2;

(k=O,

c?((O)=-w

...,m f l ) .

h r t h e r m o r e , i t follows from 'keorem 21.4 t h a t

.

C(0) # o

(23.3)

We s h a l l prove t h e f o l l o w i n g theorem. 'THEOREM 73.1: (23.7)

At

a=O

,

k(0) =-w

(k=O,

...,m t l )

arl 0

(23.4)

(k=O,

$(O)=ltw

Proof:

W e s h a l l prove t h a t

(23.5)

c(0) = l + w

...,mtl) .

.

Then, t h e proof o f Theorem 2 3 . 1 will be completed.

Observe t h a t

(23.6)

Hence ( 2 3 .?!

and

'.' This

s e c t i o n i s p a r t i s i l y based on Y. Sibuya [LO].

PROPERTIE AT

-1

(0)= ( w

W

(23.8)

091

a=O

91

- ~)~,(O,O)bl$(O,O)

Formula (21.21) implies t h a t

1

?n

w091(0)=zW4 # O .

(23.9)

Therefore, (23.5) follows from (23.61, t h e proof of Theorem 23.1.

(23.7) and (23.8).

This completes

I n c i d e n t a l l y , we have a l s o derived

1

ym(O,O)y;(O,O)

(23 .lo)

Next, we s h a l l i n v e s t i g a t e t h e d e r i v a t i v e s of

Assume t h a t

THEOREN 23.2:

I?%

and (23 -1 2)

Proof:

If

m

(23 -13)

.

= 2Wi;me1(1-w )-1#0

m23

C(a)

and

?((a)

at

a=O

.

m i s odd.

#o

(j=2,

...,m) .

a=O

j

i s , odd, formula (21.21) becomes 1

w

(a)=2w

P

,

0 9 1

and hence

1

TI-1

( 23.13 ‘ )

w1 9 2 ( a ) = 2 w 4

This means t h a t

W

192

.

(Cf. (21.19) .)

i s independent o f

(a)

c i e n t t o i n v e s t i g a t e t h e d e r i v a t i v e s of

W

a

092

.

(a)

Therefore, i t i s s u f f i -

.

set

aym( 0,a )

(23

Observe t h a t

cpjw=

aa j

Q; ( 0,a) 9

qj(a) =

aa

j

( C f . (21.15),.)

L e t us

STOKES MULTIPLIERS

92

and hence (23.16)

...,m) .

( j =1,

I n the same w a y , i t follows from (23.13) t h a t -1

(23.17)

(W

_-

- w )bi(o,o)cpj( 0 ) + (W (j=1,

j =1

If we s p e c i a l i z e (23.17) far

G2- l ) y m ( O , O ) Since m 2 3

, it

(23 .lE)

, we

...,m) .

get

q o )=0

*

follows t h a t

bm(O,O)$,(0)

=O

.

Then, by s p e c i a l i z i n g (23.16) f o r

awe;$)

j =1

(a=O = 0

, we

get

.

This proves ( 2 3 . 1 1 ) .

Then

D. =(w -1 - l ) ( l - u - ' ) ( u - j - ' J

(23.20)

-l)(w-1 - w - j ) z 0

for

j=2,

(23.21)

...,m .

This lceans t h a t , if

f o r some

j

of (23.21), then

Urn'(O,O)Cpj(O)

=o

9

ym(o,o)$j(o) = o

.

Therefore, (23.10) implies t h a t , if (23.22) holds for some

we must have

j

of (23.21),

PROPERTIES AT

( 2 32.3)

cpj(0)=O

,

a=0

93

f o r such

$.(O)=O

J

.

j

We s h a l l prove t h a t (23.23) i s impossible f o r any j

, where

j = 1,

To do this, l e t us s e t

m ~ , ( x , a ) =I)(x) +

i n t h e neighborhood of

a=O

za T

j =1 j j

( x ) + o(lall

2

, where

It is easy t o v e r i f y t h a t

where

Theref ore,

lim q ( x ) = o x-) so, Urn q j ( x ) = O x-) +o,

Furthermore,

Observe t h a t

,

l i m 7jI(x)=O X-)

,

+=

,

Urn T l ( x ) = O j

X-++

.

+ ...+

2

)

,

...,m .

9.L

STOKES MULTIPLIERS

I'hese tuo l i f f e r e n t i a l e q u a t i o n s imply t h a t d- [r r , l ( X ) y j ( x )- T ( x ) y ; ( x ) l =-X"-jy(x) 2 -dX

,

heme

an.!

i-

- r : f ( O ) Q . ( O ) =-J xm-j1,(.)2dx

TI(OjTy0)

(23.24,

J

Nsu, i t can be proved w i t h o u t

sm

J

.

0

any d i f f i c u l t y t h a t

x m - j ~ ( x ) 2 d x # 0.

is

Tn f e c i , i: cm be shown t h a t

y(x)/T(O)

i s real-valued f o r

x20

.

T h e r e f c r e , i f ' +,he right-hand member o f (23.24) i s z e r o , w e must have 7(

jo = L' f o r a l l

Tor m y

j

.

.

x>O

'I'his i s i v p o s s i b l e .

Hence (23.23) i s i m p o s s i b l e

'This proves : 2 3 . 1 2 ) , and t h e proof' o f Theorem 23.2 i s com-

pl-eted. D'cserve t h a t (23.24) can be w r i t t e n a s

anli th;t

(23.25)

cJ#-1

?'here:'ore,

(j=l,

...,m ) .

by g t i l i z i n g (23.17) and (23.24

I)

,

we can e x p r e s s t h e q u a n t i t i e s

b p , o ) c 3 j ( o ) , ldm(o,o)tj(o: i n terms o f t h e i n t e g r a l

J'

+ = .

X ~ - ~ ~ ~ (2dx X , .O )

0

The f u n c t i o n b m ( x , o )

s a t i s f i e s the d i f f e r e n t i a l equation

m

y"-x y = 0

(23.26,'

and t h e ooundary c o n d i t i o n ( 2 3 .a'

y+O

as

X-C+

.

Henze

(23.281 where is

~i

b,(x,O)

H(l'(5) P

=C~~ P H ( ~ ),( S )

i s t h e Hankel f u n c t i o n of t h e first k i n d o f o r d e r

c o n s t a n t , and

p

,

c

PROPERTIES AT

8

=0

95

The d i f f e r e n t i a l equation (23.26) has been studied by various experts .%

Let us consider the case when m 2 2 and m is even.

-c(a) =

( 2 3 -29)

-w

1-2v ( a )

In t h i s case

9

where v ( a ) =bl

(23.30)

(a)

.

( C f . (21.11) and (21.12) .)

P+l If

LEMMA 23.1:

ah=O

(23 -31)

-

Proof-

( al,

The q u a n t i t i e s

..., am )

for

h=l,

bh(a)

...,i12n , m12 + 2 ,...,m

a r e defined by (6 .7 ).

,

This means t h a t i f

s a t i s f i e s condition (23.31), we obtain 1 -j 2

abh(a)

7c

= z -

-h

[1+ Hence ( 23.32) follows.

THEOREN 2 3 . 1: Assume t h a t

YI

(23.33)

and,

m)_2

m is even.

a=O = O ’

if m 2 4 ,

#o

(23.34)

Proof:

(j=2

,...,y1 , 1p 2 ,...,m) .

m is even, we have 1 -r 3 - v (a) Wo,l(a) = 2~ m,l= 2u4 9

Observe t h a t , i f

and

*

~~

~

~

See, f o r example, C . A . Swanson and V.B.

Headley [ 4 4 ] .

96

STOKES MULTIPLIERS

and

la=O

( C f . Lemma 23.1.)

Note a l s o t h a t

W, , ( a ) C(a) =

w l' , p

-

(Cf. ( 2 1 . 1 5 ) . )

Therefore, we can prove Theorem 23.3 i n e x a c t l y t h e same way a s w e proved Theorem 23.2.

The details a r e l e f t t o t h e r e a d e r s .

The method given above does n o t y i e l d any i n f o r m a t i o n about t h e derivative of

Cia) with r e s p e c t t o

a

.

The method i s based on t h e

t h r e e formulas:

(Cf. ( 2 3 . 1 6 ) , (23.17) and ( 2 3 . 2 4 ' ) . ) become

In c a s e

j =1p l

, these

formulas

PROPERTIES AT

a=O

97

and (1111 )

respectively.

F'urthemore, i n this case, w e have

These r e s u l t s are n o t enough t o derive any conclusion about t h e d e r i v a t i v e of

C(a)

with r e s p e c t t o

.

al

P1

Let us go back t o Theorem 22.2 f o r case m = 2 t h e proof of this theorem i n Section 22.

.

W e did not complete

We must prove (22.21). To do

t h i s , set

1 C(0,2b) =&

(23.35)

erieaHP1

.

r (h)

( C f . (22.20) and (22.201).)

S p e c i a l i z i n g Theorem 23.1 f o r m = 2 , w e o b t a i n

Since

1 =A , i t r(2)

(23 -37)

follows from (23.35) and (23.36) t h a t

ep=1,

or

B=O.

Therefore,

( 23 -35

C(0,Zb) =&

1 ri cab,,, 1 . e

r($w

Py d i f f e r e n t i a t i n g both s i d e s of (23.35'1 with r e s p e c t t o we d e r i v e t h e following formula:

b

at

b=O

,

98

STOKES MULTIPLIERS

y

where

If

i s Euler’s c o n s t a n t .

a

= log 2 -

1

pi , then

(23.38) gives the a2 a t a = 0 Note

.

value of the d e r i v a t i v e of C(a) with r e s p e c t t o 1 The value given by (27.38) involves E u l e r ’ s t h a t , i n t h i s case, -2m t l = 2 constant. This i n d i c a t e s the d i f f i c u l t y of computing the d e r i v a t i v e of

.

C(a)

w i t h respect t o

at

a

a-0

i n general.

+l In order t o f i n d

a

, we

s h a l l use results obtained i n Chapter

we s p e c i a l i z e Theorem 19.1 f o r

m=2

, we

4.

get

and (23.40)

b;(x,al,a2+

A)

-1

X

A

= C 2 ( 5 , a2tX)[-1+F2(x,X) ]P(x,X)‘exp{-S

-1

2 p ( T , l ) dT1

.

0

Here,

2 ( i ) P(x,A) = x + alx+ a2 + X ; (ii) Fl(x,A) and F2(x,h) tend t o zero uniformly f o r /all + lazl I R o as

x

9

tends t o i n f i n i t y i n t h e sector:

(iii) Fl(x,X)

and

F (x,h) 2

x

(iv)

tends t o

Sa3

the q u a n t i t y

(23 -111) where (23.42)

larg A1

5.-

Po ;

tend t o zero uniformly f o r

1a-J + l a 2 \ ‘-Ro

as

OlX<+=

9

larg XI

57-

Po

9

1k1

; “C(al,a2t A )

has the form*

^c2 ( a1 ,a 2 t h ) =expCL2(al,a2+

A)

3 ,

1 1 2 -1 -(a +x) - -a += 2 1 2 2 ’JdT L2(al,a2+A) =J [ P ( T , ~ ) - T - p1T+ 1 0

L e t us s e t x=O

,

al=O,

Then

’’ C f .

2Mo

(19.21) and (19.26).

a2=0,

X=2b>O

.

.

If

PROPERTIES AT a = O

L2(0,2b)=J

*

2

[(T

-I2

t2b)

b

-T--]dT rt 1

99

,

0

and

lim F1(0,2b) = O

,

l i m F2(0,2b) = O

b++-

b-+*

.

Observe that

where l i m F(b) = O b+-b

.

Furthermore, [1p + v ( a )

I

wl,2(a) =2w

becomes

FI i ( b - i )

(0,2b)= 26 (23.44) 192 Therefore, it follows from (21.15), (23.43) and (23.44) that

W

( 23.45 )

From (23.35') and (23.451, we obtain

100

STOKES MULTIPLIERS

(23.46) A straight-forward computation y i e l d s

1

L2(0,2b) = 2 b { & t

2 t log b ) ] 4 2 l o g 2 - - 4( l o g

(23.47)

,

and hence 2L2(0,2b) = b { l t log 2 - l o g b]

(23 4 ' 7 ' )

.

Furthermore, r (1 2 t b ) =&

(23.48)

exp{b l o g (1~ t b ) - b - 21 ][lt0(b-1)]

=&(1t O(b-'))exp{b

l o g b - b)

.

Therefore, from (23.46), (23.47I ) and (23.48), w e g e t 1

a=log2-ni 2

.

This completes t h e proof of Theorem 22.2. Asymptotic behavior of Stokes multipliers a s

24.

a,,,

tends t o i n f i n i t x .

Let us s e t c(a,A) = C ( a l

(24.1)

,...,am-1'

am t h )

.

In Sections 25 and 26, w e s h a l l prove the following theorem.*

.

THEORplu.1: Assume t h a t

m23

( I ) the Stokes multiplier

c(a,x)

c(a,A) =exp{K(l-w

(24.2)

admits an asymptotic r e p r e s e n t a t i o n

lti % )A

(1t 0(1))}

uniformly f o r (24.3)

as

la11

A

...+Ia,I

Go

tends t o i n f i n i t y i n the s e c t o r

-2.

(24- 4 ) where

+

boIarg X ( 2 n - 2 -

Ro

b0

,

i s an a r b i t r a r y b u t fixed p o s i t i v e number,

6,

i s an a r b i t r a r y

but f i x e d s m a l l p o s i t i v e number and 7

7

(24.5) 0

*

Cf.

Y. Sibuya [ 3 5 ] , and H. Gollwitaer and Y. Sibuya [13].

101

(11) c ( a,X) a l s o admits an asymptotic r e p r e s e n t a t i o n 2 11 1+- -tc(a,A) =exp{K(l-w m)A2 y l + o ( l ) ) )

(24.6)

2 11 -1-2; t expEK(1-w m)A (1+0(1))] A

uniformly f o r (24.3)

tends t o i n f i n i t y i n the s e c t o r

(24.7)

The n o t a t i o n

o(1)

Therefore, (24.2) and (24.6) give us an information about the

asymptotic behavior of

25.

A

The two s e c t o r s (24.4) and (24.7) cover A-plane com-

tends t o i n f i n i t y . pletely.

denotes a q u a n t i t y which tends t o zero a s

c(a,X)

P(x,X) = x3 t A

Case when

as

.

X

tends t o i n f i n i t y i n any d i r e c t i o n .

To i l l u s t r a t e our method, we s h a l l consi-

d e r t h e d i f f e r e n t i a l equation (25.1)

ytt

i n t h i s section. (25.2)

- (2+ x ) ~ o=

Set f(x,A) =U3(x,0,0,A)

The f u n c t i o n f(x,A) perties. (i) f

.

is a s o l u t i o n of (25.1), and has t h e following pro-

(Cf. Theorems 6.1 and 19.1.) and

f

admit t h e asymptotic r e p r e s e n t a t i o n s

I

uniformly on each compact set i n A-plane a s

x

tends t o i n f i n i t y i n any

closed subsector of t h e open s e c t o r lug

(25.4) (ii) f

and

f

The constant K

I

XI


i

a l s o admit t h e asymptotic r e p r e s e n t a t i o n s

i s given by

102

STOKES MULTIPLIERS

1 2

sro

( ( t 3 t1 ) 2 -t 2 ) d t

K=J

(25.61

.

0

i5.e q u a n t i t i e s

Fl

and

F2

tend t o z e r o uniformly f o r

0 5x < b as

1

tends t o i n f i n i t y i n t h e s e c t o r

(25.7) where

la%

q

In-Po

I h l ,No

9

i s an a r b i t r a r y b u t f i x e d p o s i t i v e number, and

po

t i v e n m b e r depending on uniformly f o r ( 2 5 . 7 ) as If

9

m=3

,

x

the sector larg

(25.8)

.

po

The q u a n t i t i e s

tends t o

9

SCo

F1

and

Mo

i s a posi-

F2 tend to zero

.

i s given by

"-2k71 <: TI ,

2nd 2 w =exp(i-TI]

(25.9) The f i v e s e c t o r s each s e c t o r (25 .lo)

-

5

%

9 ,a

.

(k=0,1,2,3,4)

cover x-plane.

subdominant s o l u t i o n i s given by

y=f(w-kx,w-3ka)

.

F i g . 25.1.

( C f . F i g . 25.1.)

In

fo(x,X) =c(A)fl(x,X) - wf2(x,h)

(25.12)

Here, we used (21.11) f o r m = 3

-C(a)

=-w

.

, i.e.

.

The main problem i n this s e c t i o n i s t o derive

( C f . (21.15) and (23.13’) .)

t h e asymptotic representations of

c(X)

X

as

tends t o i n f i n i t y .

Note t h a t Wonskians (25.13) a r e independent of x (A) a t w0,2

x=O

w

.

-6 = w -1

If w e s e t

’

--1 f(0,X) = [ 1 + 0 ( l ) ] X (25.16) fl(0,X) = [ - l + o ( l ) ] A

A

(25.19)

tends t o i n f i n i t y .

x=0

i n ( 2 5 . 5 ) , we g e t

2

‘exp{Kh

61

,

2

exp {KA6]

tends t o i n f i n i t y i n s e c t o r (25.7).

hold as

W e s h a l l compute

, i.e.

where we used the i d e n t i t y

as

.

Therefore, i n t h e s e c t o r

W e can w r i t e s e c t o r (25.17) as

- y + p o < a r g XLn-po

.

(See Flg. 25.2.)

STOKES MULTIPLIERS

104

I:

A- pQane

3 --T

5

Flg. 25.2. I n this s e c t o r ,

as

tends t o i n f i n i t y .

A

Hence

-1 (25.21)

as

5 5

c(A) =w2[1+ o(l) ] e x p { ~ ( l w-6)h61 + tends t o i n f i n i t y i n (25.19)

A

.

The method used i n s e c t o r (25.19) does n o t apply o u t s i d e t h i s s e c t o r For example, l e t us c o n s i d e r a s e c t o r d e f i n e d by (25.22)

- n t p,<arg

A<-$-

po

.

(See f i g . 25.3.)

A-plane

Fig. 25.3.

105

In s e c t o r (25.22), we have larg

(25.23)

AI

5n-

Po >

I.rg[w%lt

n p0-

~

Hence, (25.16) and

--1

I

=ft(O,wl*A) =[-l+o(l)]wA4 exp[Kw 3 X6 ]

f'(0,w''A)

hold as

A

222

=f(0,;4h) =[It o ( l > ~ w -4 ~exp[Kw ~ 3 61 , -1 lo5

f(O,w-'A)

( 2 5 .Y)

a

If (25.16) and (25.24)

tends t o i n f i n i t y i n s e c t o r (25.22).

are i n s e r t e d i n t o (25.15), we g e t

lo2 (1) = o ( l ) e x p [ K ( l + w 3 )X 6 1 012

W

This estimate is not s a t i s f a c t o r y .

.

In order t o get a b e t t e r estimate

we

s h a l l use t h e following f o e a s :

(25- 2 5 )

I

f(0,A)

= 1 7 [ ~ ( o , ~+~wIf ()0 , w 2 uI C(W

f '(0,A) =-

A>

I

[wf l(o,W%)

+ fl(0,w2A) I ,

c(w3A)

which are derived from (25.12) by replacing A

(25.25) i n t o (25.151, we get (25.26) where A(k) =

(25.27)

and B(A) =

(25.28)

It i s n o t d i f f i c u l t t o prove t h a t A ( A ) =wu0,~(w3a) ,

(25.29) To estimate

A(?)

, we

assume t h a t

3

by w X

.

Upon i n s e r t i n g

STOKES MULTIPLIERS

-

A plane

In s e c t o r (25.30) we have l=g[w 3A J (~T I -

(25.31)

Po

1arg[w%11

P,

.

Hence, (25.24) and

-2 -1 52 f(0,w3X) = [ l + o ( l ) ] w % exp[Xw2X6] , 21

(25.32) €[(Q,w3X)= [-l+ o(l) hold as

A

25

1 ~ 5 exp[h2X61 ,~

tends t o i n f i n i t y i n s e c t o r (25.30).

If (25.24) and (25.32)

a r e i n s e r t e d i n t o (25.27), we g e t

g

2 (25.33)

as

A

A(h) = 2w4[l -k o(l) ]exp[X(w tends t o i n f i n i t y i n Sector (25.30).

3 s e c t o r (25.30), the q u a n t i t y w X

- l)h

as

A

2 1

tends t o i n f i n i t y i n s e c t o r (25.30).

1 ( A )= 2 W 4 [ [ l t d

092

or

13

is i n s e c t o r (25.19). Hence

c(w3X) = w 2 [ l t o ( l ) ] e x p [ - X ( l t w

W

1

Wthermore, i f

-1 (25.34)

s6

-6

6 )A ]

Therefore,

lo o(1) ]exp[K(w

5 2

-6

+ w )A

2 1 -6 6 + [ l + o ( l ) ] e x p [ K ( l + w ) A 11

6

1

is i n

CASE WHEN

P(x,A) = x 3 + h

-

i 2 5

1 c(A) = w z { [ l +

(25.35)

107

-6

o(l)]exp[K(w

- w6 ) A 6 1 1 2

-6

+ [ 1 + o ( l ) I e x p [ K ( 1 + ~ )A as

tends t o i n f i n i t y i n s e c t o r (25 .30)

A

arg A

The neighborhood of the d i r e c t i o n s e c t o r s (25.19) and (25.30).

.

6

11

=-$

I n order t o estimate

is n o t covered by i n the neigh-

(X)

W 092

borhood of t h i s d i r e c t i o n we use the following formulas:

(25.36)

I

= 1 2 [ f (0,w2A) + w f ( O,wh)

f(O,w5)

J

,

c(w A ) f ‘(0,w-lA)

=y [wf ‘(0,w2A) + f ‘(0,WA) 1 , 4 w

A)

which a r e derived from (25.25) by replacing

A

by

w-~A

.

Upon i n s e r t i n g

(25.36) i n t o (25.15) we g e t

I n the same way a s we derived ( 2 5 . 2 9 ) , w e g e t

..

2

(25.39)

= L 4,

In order t o estimate

@A)

( 25 -40)

B(A) = W W ~ , ~ ( A )

, we

-TT+p o < m g

-7

..

.

assume t h a t

n

X i - 5-

po

.

( C f . Fig. 25.5.)

d(25.40)

~

c -

A- Plane Fig. 25.5.

STOKES MULTIPLIERS

108

In s e c t o r (25.4O), w e have (25 -41)

lmg

Ln-

P,

IardwAIl

9

5.- P,

-

Hence, (25.16) and

52. 11 % 4 exp[& 6x 6 I ,

I

f(0,wX) = [I+o ( 1 ) 1w

(25-42)

11

fl(o,wa) = [-i+ o ( i ) 1w%4

hold a s

tends t o i n f i n i t y i n (25.40)

X

.

Ii 6 6 exp[Kw a I , Thus

i.z

2

B(A)

(25.43)

6 6

3

= 2 W 4 [ l t o(l)]exp[K(ltw ) X

as

tends t o i n f i n i t y i n s e c t o r (25.40). Furthermore, i f 2 (25.4O), the q u a n t i t y w X i s i n (25.19). Hence c

1

as

5

tends t o i n f i n i t y i n (25.40)

.

z

2 5 (1tw6)w3=-(1+w-6)

(25.45)

A

is in

c c

Observe t h a t

.

P u t t i n g everything together, we g e t again (25.35) a s

X

tends t o i n f i n i t y

.

i n (25.40)

Thus we obtain (25.21) as

1

-

1 2 1 c(X) =w2[1+o(l)]exp[K(1+w-6 ) A 6 ] tends t o i n f i n i t y i n the s e c t o r

c

-2

+ as

A

c

6 6 [ 1 t o ( l ) ] e x p [ K ( 1 t w ) h 13

tends t o i n f i n i t y i n the s e c t o r

(25.46) (See Fig. 25.6.)

-n-$+Po<arg

1s:-

p,

.

Sectors (25.19) and (25.46) cover a-plane completely.

3

CASE WHEN P ( x , h ) = x + h

log

Mg. 25.6. Now, we s h a l l investigate t h e r e l a t i o n between the two asymptotic formulas (25.21) and (25.35) i n the common p a r t of the two s e c t o r s (25.19) and

(25.46).

To do this, l e t us f i n d the term i n formula (25.35) which domiSuch a term can be found i f w e examine t h e real p a r t of

nates t h e o t h e r . the quantity

2 2

(25.47)

2 2

2 2 2

‘6 6 6 K ( l + w 6 ) A 6 = K ( ~ + w - ~ ) A ~ - K (- ww ) A

i n s e c t o r (25.46).

Observe t h a t

I 6

(25-48)

arg(l+w )

= 6n

.

Hence, i n (25.46), (25.49) This implies that

F

STOKES MULTIPLIERS

110

i n s e c t o r (25 .46).

Therefore, if b 0 > 0

i s g i v e n , we have

-1 (25.51)

c(h) =w

2

2

5 2

-6 6 6 [ l + o ( l ) ] e x p [ K ( w - w )A ]

and

1 2 5 -6 6 c(A) = w 2 [ 1 t o(l)JexpCK(1 t w ) A 3

(25.52)

as

tends t o i n f i n i t y i n t h e s e c t o r

A

(25.53)

-r-+n+

p 0 < a r g 11-9tio

and i n the s e c t o r

-4+ 5 t i o ~ a r g1515- p, ,

(25.54) respectively.

(See F i g . 25.7.)

The common p a r t o f s e c t o r s (25.19) and

(25.46) i s contained i n s e c t o r ( 2 5 . 5 4 ) .

Thus w e found t h e r e l a t i o n between

t h e twc asymptotic formulas (25.21) and (25.35) i n t h e common p a r t o f

(25.19) and ( 2 5 . 4 6 ) . T

Mg. 25.7. Now, l e t u s make t h e f o l l o w i n g o b s e r v a t i o n .

(25.55)

-+ n

p,<arg

A52n-$-

6,

Assume t h a t

.

5

Then - n - $ n t p o 5 a r g [ w - 5 ~ ~ < - $ - 6, and hence t h e q u a n t i t y

wV5A

-1

,

i s i n s e c t o r (25.53).

5

Therefore

2-32

c ( h ) =w2[1+o(l)]exp[K(w -6 - w 6 ) w 6 k 6 ]

as

1

t e n d s t o i n f i n i t y i n sector (25.55).

Since

CASE WHEN

and

2-3

6 6 =w w w we g e t again (25.56) as

A

-? =w

P(x,X) = x 3 + A

111

-:-;2 , =-w

w

-6

-

1 5 2 c(A) = w 2 [ l + o ( l ) ] e x p [ K ( 1 + w-6 ) A 6 1 tends t o i n f i n i t y i n (25.55)

.

Thus we have proved the following

result: The q u a n t i t p

admits an asymptotic representation

c(A)

-1 2 2 c(A) = w 2 [ l + o ( l ) ] e x p [ K ( l + w-6 ) A 6 1 as

A

tends to i n f i n i t y i n t h e s e c t o r

-?+b 0 5 a r g A ( 2 n - y -

b0

.

Furthermore, asymptotic formula (25.35) i s v a l i d a s

?,

tends t o

i n f i n i t y i n the s e c t o r (25.57)

IargA+?l
Therefore, i n the s e c t o r (25.58)

larg A

- 2n+$-J IS, ,

we have (25.59)

-1

5 2

c(X) = w 2 { [ 1 +

o(l)]exp[K(l+ w-6 ) h6 1

5 2 + [ 1 + o ( l ) ] e x p [ K ( 1 + w6 )A 6 1)

as

X

tends t o i n f i n i t y .

is i n s e c t o r (25.57).

Note t h a t , i f

X

i s i n s e c t o r (25.58),

Thus, by u t i l i z i n g formula (25.35) f o r

c(w'~A) , we g e t (25.59). 24.1 f o r P(x,X) = x 3 + A

.

=

By using these results, we can prove Theorem

I n this s e c t i o n , we used formulas such as

f ( 0,A) =

c(X)

LJ%

3c ~ ( o , ~+wf(0,w2A)) ~A) , c(w A )

STOKES m T I P L I E R S

112

These formulas are analogous t o a well-known formula f o r t h e gamma function: n

= sin(nA)

r(h)r(i- k )

(25.60)

The asymptotic r e p r e s e n t a t i o n of

.

r(h) :

which i s known as S t i r l i n g ' s formula i s v a l i d as

A

tends t o i n f i n i t y i n

the s e c t o r

.

Jarg A/ < n

Formula (25.60) y i e l d s t h e asymptotic r e p r e s e n t a t i o n of

T(1)

i n the

sector

I a r g - nl IPo As a matter of f a c t , we were n a t u r a l l y l e d t o formula

t o compute t h e Stokes m u l t i p l i e r s o f

(25.60) when we t r i e d

i n Section 22.

y,(x,al,a2)

(Cf.

(22.18) and ( 2 2 . 1 8 ' ) .)

If we examine the method of this s e c t i o n , we n o t i c e t h a t t h e estimate of

e s t i m a t e s of Wo,3jh)

26.

( A ) , while t h e 092 i n (25.30) and (25.40) are based on t h e estimates o f

i n s e c t o r (5.19) i s based on t h e e s t i m a t e of

c(X)

c(A)

W

*

General c a s e .

In this s e c t i o n , we s h a l l consider t h e d i f f e r e n t i a l

equation y " - P(x,X)y= 0

(26 -1)

,

where

m

P(x,h) =x

m23

We assume t h a t (26.2)

.

+ alx m-1

+

...+

f(x,a,k)

+a)

.

is a s o l u t i o n of (26.1), and has t h e following

p r o p e r t i e s .*

(i) f

' Cf.

a d

f'

.

Set

f ( x , a , A ) =hm(x,a1, .-.,am-1ya

The function

+ am+ X

admit t h e asymptotic r e p r e s e n t a t i o n s

Theorems 6.1 and 19.1.

GENERAL CASE

113

--1

r

f ( x , a , i ) = x m[l+ O(x 2)]exp{-Em(x,a)}

--12

(26.3)

+rra

f '(x,a,A) = x

uniformly on each compact set i n

,

[-1+O(x ) ]exp{-Em(x,a) I (a,h)

,

- space

a s x tends t o i n f i n i t y i n v- u 8-1 8, g1 ;

any closed subsector of t h e open s e c t o r

-1

(26 - 4 )

-12

X

.

I f i ( x , a , h ) = ^ C ( a , h ) [ - l t F2(x,A)]P(x,X)4 exp[-s P ( T , 1 $ d ~ ] 0

The q u a n t i t i e s

Fl(x,X) lal[ t

and

F2(x,A)

...+lam[I

tend t o zero uniformly f o r

R ~,

o<x<+=

Ia r g

h tends t o i n f i n i t y i n the sector: t i t i e s a l s o tend t o zero uniformly f o r

as

as

x

tends t o

Sa,

.

A1

5.-

po

.

These two quan-

The q u a n t i t y ^c(a,X) admits t h e asymptotic repre-

senta ti on 11

2-G( 1 + o ( 1 ) ) I Ic(a,X) =expCn

(26.5) uniformly f o r

lyl+

(26.6) as

* a * +

bmIso I a r g X I In -

tends t o i n f i n i t y i n the sector:

?,

m 2 3 , the

given by ( 2 4 . 5 ) .

Since

pendent of

am

The p o s i t i v e numbers Ro

f i x e d , and

Mo

.

sectors

% ,a

Em(x,a) and po

and po . -go,Ro...,-gmtl cover x-plane

K

is

rm are inde-

are a r b i t r a r y but completely.

I n each

subdominant s o l u t i o n of (26.1) i s given by

where we used t h e i d e n t i t y w -*=wa ( 2 6 . 6 ) , then

and

, where

depends on

The m + 2 sector

quantities

po

k

G (a)

.

Note t h a t , i f

i s a l s o i n domain (26.6)

.

a

i s i n domain

STOKES MULTIPLIERS

14

Set

wj ,,(.,A)

(26.8) We s h a l l compute

.

%,o(a,h)

W

(26.9)

I;

=

f . (x,a,A)

fk(x,a,h)

f!(x,a,h)

f$x,a,h)

It follows from (21.21) t h a t

( a , b ) =-2w d a )

9

190

where

.!p

p(a)

,

(m: odd)

1

(26 . l o )

I-

;;“-bl

(m: even)

(a)

P1 Rext, denote by (26.11)

.

To t h e sector i n A-plane which i s defined by n-z8 ntpo(arg A2n-p

.

0

I t i s easy t o s e e t h a t

A E To i f and only i f

(26.12)

I n - Po

J a r g hl

Hence i n

To

I ~ g [ w 2 - ” h l lc_n- Po

9

we can use asymptotic formulas (264 ) . By s e t t i n g

x = 0 in

(26.4) , we get

--1 (26.13)

i

fo(O,a,A) = [ I +O

( I ) J ~ ‘t(a,~)

,

-1

f;(o,a,A) = [-1-to ( l ) I ~ ~ ^ C ( ~, , X )

uniformly f o r (26.6) a s

X

tends t o i n f i n i t y i n

To

.

f 2 ( x 7 a , h )= f(w-‘x,C2(a) ,w4b) = f ( w - 2 x , C

Also, since 2

,

( a ) ,w-*‘A)

ue obtHin

(26 .U)

I

--1

L-L f2(0,a,X) = w 4

2[1+ o ( l ) ] h ‘t((c2(a)

1.

-2-* 1 1 f;(O,a,A)

uniformly for (26.6) a s

2[-1+ o ( 1 ) ]A4t(G2(a) ,w2-y),

=w X

,u2-”h) ,

tends t o i n f i n i t y i n

To

.

Therefore, in

we g e t

1 1

3--

W as

h

290

(a,A) =-2w4

tends t o i n f i n i t y .

2[1to ( l ) ] t ( a , k ) t ( G 2 ( a ) ,w2‘%)

Hence (26.5) implies t h a t

2 (26.15)

W270(a,h)= e x p [ K ( l - w

1 1

)A 2-G(1+0(1))1

,

To

,

115

GENERAL CASE

A

uniformly f o r (26.6) a s identity:

w

, where

To

tends t o i n f i n i t y i n

we used an

(2-m)(+) 1$ 2 m =-w

I n general, denote by

w

T k = {A;

(26.16) I n the s e c t o r

Tk

(26.17)

, we

the s e c t o r i n A-plane which i s defined by

Tk 2k

AETo)

.

have

.

n ]-~Po ~ a r p [ w ~ + ~ - mI X

Since

k 2k -Ic k 2(k-2)+2-m~) = f ( w k x , ~(a),w A ) = f ( w X , G (a),w Y

f,(x,a,A)

we get

--1

+

fk(O,a,A) = w &(m+2-2k) 4 [ 1 o( 1)]A

1

(26.18) fi(o,a,A) = w uniformly f o r (26.6) a s

X

-k--(mt2-2k)

4

E( Gk(a) , w 2(k-2)+2-y) -1

9

4- k 2(k-2)+2-\) [-1t o ( 1 ) ]A C ( G ( a ) ,w

tends t o i n f i n i t y i n

.

Tk-2

9

Hence i f

l a r g X I I n - Po

(26.19) i n a subsector of

Tk-2

, we

get (k-l)(lt:)

(26.20)

)A

%,o(a,A) = e x p [ K ( l - w

uniformly f o r (26.6) a s

X

(1+0(1))3

tends t o i n f i n i t y i n t h i s subsector o f

Tk-2

,

where we used an i d e n t i t y :

Observe t h a t condition (26.16) i s equivalent t o n - 8 S - 2 .

po<arg

aLn-2-

Therefore condition (26.19) i s s a t i s f i e d i n

- n < n - n8- m < n - $ ( n m+2 mt2

This means t h a t (26.19) i s s a t i s f i e d i n

Tk

.

po

Tk i f and only i f

.

for

(m: odd) (26.21)

(m: evsn)

,

.

i s even, s e c t o r (26.19) i s completely covered by those s e c t o r s Tk

If

m

If

m i s odd, l e t u s denote by

-T

t h e s e c t o r which i s defined by

.

STOKES MULTIPLIERS

116

(26.22) The s e c t o r

-T

i s a l s o given by -n+ p o ( a r g

(26 2 2 1 )

Then

\in---$+2 m l

.

1 (k = 0,1,..,$m-3))

yk

cover s e c t o r (26.19) completely.

and

m = 3 , the sector

Ncte t h a t , i f

.

po

is given by

T

-n+ po<arg A L E -

5

and this is t h e same a s ( 2 5 . 4 0 ) .

( C f . Fig. 25.5.)

Thus we proved the lemma below. &M )! A

(I)

26.1: J .

.

CL( a) ( a , h ) =-2w

W,

where -

m>_3

Assume t h a t

9 0

even)

(m:

A

uniformly f o r (26.6) =

{

.

(111)

if

+2

2,3,. .,fm-3) 1 293, *

Tk -

tends t o i n f i n i t y i n

*

-,?+

1

m is odd, (11) holds i n

-T

;

, where

(m:

odd) ,

(m:

even)

1 k=$m+3)

; a

The s e c t o r

is completely covered bx

1To,T1,. . .,T1

7-1

respectively. Let us set

if -

m

i s even

,

117

GENERAL CASE

(26 2 5 )

fo(x,a,x) = c(a,\)fl(x,a,X)

The q u a n t i t i e s solution and

(26 2 6 ) Theref o r e ,

Note t h a t (26.28) Hence

Since

we g e t

or (26 - 2 9 )

where

and

c"(a,A)

.

a r e the Stokes m u l t i p l i e r s of the

f o Kith r e s p e c t t o the two l i n e a r l y independent s o l u t i o n s

f2.

(26 2 7 )

c(a,A)

+ E(a,b)f2(x,a,\)

.a),i t

From (26

follows t h a t

fl

STOKES MULTIPLIERS

118

;(a) =

(26.30)

i0 bl

,

(m: odd) (m: even)

(a)

,

F+l

This r e s u l t is the same as (21.11). Furthermore,

(26.31)

c ( a , a ) =C(al,

...,am-1’am + I ) .

Therefore, i n order t o prove Theorem 24.1, we s h a l l compute t h e matrix

From the i d e n t i t y

i t follows t h a t

Then,

(26.36)

Let

US

write ( 2 6 . X ) as

Then, Lemma 26.1 implies t h a t 2 Nk(a,A) =exp[K(l-w

(26.37)

uniformly f o r (26.6) as

h

2,3,. (26.38) k = ( 2,3,.

11 ) X 2; ( l + o ( l ) ) ]

tends t o i n f i n i t y i n

1 ..,$m-3)

+2

k E Tk-2

, we

(m: odd)

1 1 . .,?+

Note t h a t formula (26.37) i s independent of d e r i v e from (26 .?7)

Tk-2

(m: even) k

.

Since

,if ,

. 2

w X

ET

k-3

for

119

GENEIW, CASE

A

uniformly f o r (26.6) a s

tends t o i n f i n i t y i n

.

Tk-2

Formulas (26.37) and (26.35) y i e l d 2

(26 -40)

in

c(a,A) =exp[K(l-

.

To

11

%) A %(1+o ( 1 ) ) 1 w

Also, formulas (XI.%), (26.37) and (26.39) y i e l d

2 11 (26 4)

c(a,A) =exp[K(1- w

)A

2+ t exp[K(w

5%

(1+o ( 1 ) )

l< -w

)A

2

1

(l+o(l))]

for k 2 3 and s a t i s f y i n g (26.38). Formula (26.41) is ? when m is odd. Let ua consider a small s e c t o r 0 given by

in %,2 valid i n

l w h-nl

(26.42)

For A € 3

, we

get

A

E

.


~ ~ ‘ 3 E3 To and

W

for

also

Hence i n

(G‘’(a) , w - ~ A ) f 0 092 we can use (26.33) with k = -1

(a,A) =wW

-1,l

i.l

.

Thus we g e t

Lemma 26.1 y i e l d s

and

in

fJ

.

Furthermore, we can prove without any d i f f i c u l t y t h a t

W in 0

, although

(G-’(a),w 093

-

2 -1--

2h)=exp[K(w

m

-w

2 3 11 1+. m 2m )A

this i s not a consequence of Lemma 26.1.

(l+o(l))]

Thus, in 8

,

STOKES MULTIPLIERS

120

11

2

c(a,h) =exp[K(l-

( 26 44)

+

:')i'E(l+

o(1))j

2 311 -I--

exp[K(l-w

?A2

.

m(l+o(l))]

A complete formula f o r t h e aymptotic behavior of

c(a,h)

as

A

tends t o

i n f i n i t y i s given by (26.4O), (26.41) and (26 .u). If we can determine t h e dominant term i n each of them, w e can complete t h e proof of Theorem 24.1. Let us consider f i r s t t h e s e c t o r i n h-plane given by

(26 4 5 )

-&t me2

.

6 , I a r g h l n - po

Since

for

m23

small.

, the

sector

To

i s contained i n (26.45) i f

P a r t ( I ) of Theorem 24.1 h o l d s i n

6o

To , ( C f . (26.40) .)

remaining p a r t of s e c t o r (26.45) which i s n o t covered by (26.41) h o l d s .

i s sufficiently

To

In the

, formula

Let us p u t A=(w

m

-l)h

.

Then 2 4 a r g ~ = a r g [m ~- I ] +mt2 = arg

1 2 mt2 =7+++arg m 2 m

=1T t[2v tmt2 -

h

arg A ]

.

(See f i g . 26.1.)

m 2 . m

5L

FpW 2+

'W

/

Fig. 26.1.

m ...

GENERAL CASE

121

Observe t h a t

8 6 , i a r g h i n - z t po

-mt2 I t

,

i n the p a r t of s e c t o r (26.45) outside t h e s e c t o r

To

.

Hence

This means t h a t

1 mt2 ~ t - 6 2m

l? mt2 < a r g A l n - m t - 2mpo

0-

'

and hence

Re[A]< 0

.

Therefore, ( I ) of Theorem 24.1 holds i n s e c t o r ( 2 6 . 4 5 ) . Let us consider s e c t o r ( 2 6 . 4 2 ) .

In t h i s s e c t o r , we have (26.44). Let

us put A=[w

=1 F 2t n2- n + asr g m

m

2

.

~ =1? + F a r g A

m

This means t h a t Re[ A ] i n (26 .42)

.

<0

Hence, ( I ) o f Theorem 24.1 holds i n (26.42)

.

Now, l e t us consider t h e s e c t o r defined by

(26 -46)

6,

n t po<arg A(2n-&-

.

I n s e c t o r (26.46),

(26.47)

-n

t po 5 arg[w-(mt2)A

3 < -&n

- 6,

.

Therefore, we derive from (26 41) c(a,A) = c ( a , w- ( m t Z )

= exp[K(l- w

m

-1 m t 2 ) 2 1 4 )w

2n 2 , m

(1+o ( 1 ) )

1

1+2 (mt2)2 A+& m m-w )w an x2 Y1.r 41)) 1

2 4

+exp[K(w

-1-2 m L& 2m =exp[K(l-u )h ( l t o ( l ) ) ] t exp[K(l-w

2 1 1

7%

)h

(lt0(1))].

STOKES MULTIPLIERS

122

Here we used an i d e n t i t y :

If we s e t again 2 1 1 -1--+-

l$ A=[w

m]i2

-w

,

then

i n ( 2 6 . 4 6 ) . Hence Re[Al< 0 i n ( 2 6 . 4 6 ) . This means t h a t (I) of Theorem 24.1 h o l d s i n s e c t o r (26.46). Thus we prove? (I) o f Theorem 24.1 i n s e c t o r ( 2 4 . 4 ) .

We can prove (11) of

Theorem 24.1 i n s e c t o r ( 2 4 . 7 ) , i f we u s e (26.41) and the f a c t t h a t

1 a r g [ w-(mt2)h

] - $1

5 60

ir. (2,!+.?). This completes t h e proof of Theorem 24.1. 27.

A f u n c t i o n a l e q u a t i o n . % I n S e c t i o n 25 , w e s t u d i e d t h e s o l u t i o n

(27 -1)

f l x , h ) =u3(x,o,o,A)

of the d i f f e r e n t i a l equation

y"- ( x3 t A)y=O

( 2'7 . 2 )

.

In p a r t i c u l a r , we d e r i v e d t h e asymptotic r e p r e s e n t a t i o n s (25.56) and (25.59) of t h e Stokes m u l t i p l i e r

[

C(A)

(27.3) Then (27.41

S(X) = S(,A)

-w

.

c(X)

a3

Set

'

s a t i s f i e s the condition

6

2

S(LJ%)S(W X)S(w4h)S(w h ) S ( X ) =12

'iWs i s a s p e c i a l c a s e of ( 2 1 . 3 1 ) . ug=u-2

,

w6 = w

Since

,

u5 = 1

. i n this c a s e , we have

w4zw-1 .

A s t r a i g h t - f o r w a r d computation shows t h a t (27.4) i s e q u i v a l e n t t o

*

This s e c t i o n i s based on Y. Sibuya and R.H.

Cameron

[@I.

P

v

n

0

I

0

II

x

3

P

3 PJ

P.

c

0

c+

rn

2a

c+

Y

v

-

h

P

P

h

5

-1

03 v

N 4

-

Y.

c+

0

u

.

h

.6

v

w

&

% N

h

rt

Y

v

P a

3

&I P

0

n

v

Y

2IE a

h

(D

5

0

v

m

N 4

h

P

w

5

m

m

P.

v

Y

0

Y

P

0,

m

E

v

h

0

II &

v

+

2

V

7

V

>

N 4

h

5. Y

V

03

N 4

h

a

€

N

N

Y

v

Y

N

E

h

0

v

7

E

h

E,

II

u

I E

v

Y

&

h

0

v

Y

P

El

h

0

r---l U

I E

v

Y

0 ,--.

Y

I

N

w

€

h

0

-

1

P

N

W

STOKES MULTIPLIERS

124

?he two c o n s t a n t s t h e o t h e r hand,

cp(A)

w t

-1

2 -2 w tw

and

LJ

On

are s o l u t i o n s of (27.8).

d e f i n e d by (27.7) i s a n o n - t r i v i a l e n t i r e s o l u t i o n

I t should be n o t i c e d t h a t (27.8) does n o t have any n o n - t r i v i a l

of ( 2 7 . 8 ) .

polynomial s o l u t i o n s .

Furthermore, a s t r a i g h t - f o r w a r d computation shows

t h a t i t i s impossible t o determine a n o n - t r i v i a l e n t i r e s o l u t i o n as a power Note t h a t (27.8) i n v o l v e s two terms

s e r i e s s a t i s f y i n g (27.8).

.

~p(u-’A)

cp(wh)

This means t h a t (27.8) i s an e q u a t i o n of t h e second o r d e r .

and

In

o t h e r words, w e can e x p e c t t h e e x i s t e n c e of a s o l u t i o n o f (27.8) which c o c t a i n s two a r b i t r a r y q u a n t i t i e s . L e t us w r i t e t h e e n t i r e f u n c t i o n

d e f i n e d by (27.7) i n t h e form

cp(k)

c p ( A ) = G o ( z ) t XG1(z) + A 2G2(z) t X 3G 3 ( z ) tA4G4(z)

(27 -9)

where t h e q u a n t i t i e s

are entire i n

Gj(z)

z

=A5

.

,

,

S i n c e cp(0) = G o ( 0 )

we have

i27.1C)

G o ( 0 ) =w2 t w -2

.

Set 2 -2 H ( z ) = 1 t (w t w ) G 0 ( z )

(27.11)

is an e n t i r e f u n c t i o n of

. which i s n o t i d e n t i c a l l y equal t o

,Then,

K(z)

zero.

Let us prove t h e f o l l o w i n g theorem.

THEOREM

27.1:

The f u n c t i o n

z

d e f i n e d bx (27.7) admits a r e p r e s e n t a -

cp(h)

tion (27.12)

cp(~)

= a l l + (1t a

2

)

[ w2S( A)tw

-2

~ ( A ) T ~ ( A -)

v( A)-1 ][ w-‘5(

B2 3

9

A)tw2q( h ) -11

where ( 2 7 J3)

a = w t w

-1

P = W2t

,

w

-2

Proof: Let (27.15)

where that

2)

X.(z) J

are entire i n

z

and

z =A5

t n + - ) + a 2 x p t 2 x p ++x4(z) , . Then equation (27.8) i m p l i e s

A HJNCTIONAL EQUATION

By computing

Xo,X1,X2,X3

and

X

4

125

i n terms of Go,G1,GZ,G3 and G

4 we

obtain

(27'16-0)

Go(a)+Go(z)2t

~sG,(z)G

(a) +azGZ(z)G 3(z) = 1

4

,

and

If we i n s e r t (27.17) and (27.18) i n t o

(27.19) we g e t

and hence

(27.16-O) , and

use t h e i d e n t i t y 2 2 1-Go( z)-Go( z )=~(l+aGo( a)) (1+ pG0( z)) = H( z)(l+a -a H( z ) )

,

STOKES MULTIPLTERS

126

= -(:+I)-

l)(w5tw-111-l)(w

-1t

+wV-l)

. .

'vle ( c a e a s i l y d e r i v e ( 2 7 . 1 2 ) from ( 2 7 . 2 0 ) , ( 2 7 . 2 1 ) , (27.22) and (27.23) ?'his completes t h e proo:' ue dic, n o t u s e

that

M(z)

of Theorem 27.1.

(27.16-1) and (27.16-4).

In t h e proof o f Theorem 27.1, A c t u a l l y , by u t i l i z i n g t h e f a c t

i s n o t i d e n t i c a l l y e q u a l t o z e r o , w e can d e r i v e t h e s e two

r e l a t i o n s from (27.16-0), (27.16-2) and (27.16-3).

I t i s e a s y t o show t h a t t h e right-hand member of (27.12) r e p r e s e n t s a solut3.on of' ( 2 7 . 8 ) which c o n t a i n s two a r b i t r a r y q u a n t i t i e s

.

X-4T)(A)

and

This means t h a t (27.8) admits n o n - t r i v i a l r a t i o n a l s o l u t i o n s

and n o n - t r i v i a l meromorphic s o l u t i o n s . s u i t a b l e way o f choosing entire in

l-'S(k)

h

.

A-'S(X)

and

However, we have n o t found any A-%(h)

so t h a t

cp(h)

becomes

The e x i s t e n c e of a n o n - t r i v i a l e n t i r e s o l u t i o n of (27.8) i s

known, 'cut it, is s t i l l n o t c l e a r why (27.8) admits such a s o l u t i o n .

:1 i s known t h a t t h e gamma f u n c t i o n i s t r a n s c e n d e n t a l l y transcendental.

In o t h e r words, t h e gamma f u n c t i o n does n o t s a t i s f y any a l g e b r a i c d i f f e r e n t i a l e q u a t i o n s . This f a c t can be proved by u s i n g t h e d i f f e r e n c e equation:

r(At 1)= U ( k )

d e n t a l l y transcendental? either.

only."

Is t h e f u n c t i o n c p ( A )

a l s o transcen-

Me do n o t have any answer t o this q u e s t i o n ,

I t should be n o t i c e d t h a t we can n o t prove o r d i s p r o v e t h e t r a n -

scenden t a l transcendency of rp( x) by u s i n g (27.8) only, s i n c e ( 2 7 . 8 ) =hits non-trivial r a t i o n a l solutions. By s p e c i a l i z i n g i d e n t i t y (21.31) for m = 3

i. l i e b e r b a c h [3; pp 325-3281.

, we

can d e r i v e

A FIINCTIONAL EQUATION

(27.24)

NS

,TI) + N U S

,W-T’ I)

~ ,q) S= 1

F

127

,

where (27.25)

F(S,T]) =w2C(0,S,TI)

-

If we regard t h e right-hand member o f ( 2 7 . 1 2 ) as a f u n c t i o n of

this r a t i o n a l f u n c t i o n s a t i s f i e s equation ( 2 7 2.4)

.

and

‘Tl

,

CHAPTER 6 A BOUNDARY VALUE PROBLEM I N THE COMPIXX PLANE

28.

Soundary conditions.

L e t us consider an eigenvalue problem

m m-1 y - [ x t a 1x (Pjk)

(x-t-

YE0

+ ...t a m ] y = h y , i n % and i n 8.) J

i n x-plane, where 2kll s: [ a r g x--[m-2
sj: l a r g In c e r t a i n cases problem

x - 2 1

<&.

(P. )

becomes t r i v i a l .

Jk

For example, if k = j

,

all values of h are eigenvalues, w h i l e t h e r e are no eigenvalues i f k =j t 1 I n t h i s chapter, w e s h a l l study problem ( P ) In o t h e r

.

092

.

words, we s h a l l study t h e eigenvalue problem

(28.1;

y”-P(x)y=Xy

,

YE5

,

where (28.2)

and

m m-l P ( x ) = x t a1x t

...+am ,

5 i s t h e set o f a l l e n t i r e functions o f x which s a t i s f y t h e condi-

tions

628.3) as

*

x

ymo tends t o i n f i n i t y i n t h e two s e c t o r s

This chapter is based on Y. Slbuya [ 3 8 ] .

LARGE EIGENVALUES

s2: 1 a r g x -&+ <& I.

(28.5)

129

(See Fig. 28.1.)

Fig. 28.1. I n this chapter, we a s s m e t h a t

al,

I n the case m = 1 , the s e c t o r s

...,am-

a r e given constants.

go , S1

s2

and

cover x-plane com-

p l e t e l y , and there i s no s o l u t i o n which i s subdominant i n Therefore t h e r e i s no eigenvalue of problem (28.1) the s e c t o r s

- - So,

al, 8,

and

a3

.

So

I n the case

cover x-plane, and i f

y

S2

.

m=2

,

and

i s an eigen-

function of problem (28.1), we must have (28.6)

x

as

1 2 1 y(x)expEp + T ~

X I=OW

tends t o i n f i n i t y i n any d i r e c t i o n , where r=-L+k2-L(a + I ) 2 8 1 2 2

.

Since the left-hand m e m b e r of (28.6) i s e n t i r e i n y ( x ) = [ a polynomial i n In f a c t , these polynomials i n Furthermore, the eigenvalues

x

A

xlexp[-Y

x

, we

must have

\

- kia1x]

.

are given i n terms of Hermite polynomials.

must s a t i s f y t h e condition

-&+ L2 &(a + 1) = a 2 81-2 2

non-negative i n t e g e r .

Completeness of eigenfunctions and an expansion theorem i n terms of eigenfunctions are well-known i n this case.* problem (28.1) f o r

29.

*

m23

I n this chapter, we s h a l l study

.

D i s t r i b u t i o n of l a m e eigenvalues.

As w e d i d i n Section 26, l e t us

See, f o r example, E.C. Titchmarsh [45; Chapter IV, gd.2, pp 73-75].

A BOUNDARY VALUE PROBLEM

130

set

and i ,212 -1)

where

,w

-2k

a2,

.. .,w -mk am)

and

2 ~ : = e x p [ i - - n ]. mt2 Then

f2 a r e subdominant s o l u t i o n s o f

and

f',,

y" - P(x)y = xy

i 29.2) in

8,

and

s2 ,

respectively.

( 2 8 . 1 ) a r e z e r o s of t h e Wronskian

T h e r e f o r e , t h e e i g e n v a l u e s of problem

,

W (a,X) 092

where

(29.3) Ey v i r t u e of Theoren 24.1,

l-el

i29.4)

W

092

(a,A) = e x p [ K ( l - w

1+2 m 2m )A

(1+0(1))1

uniforrny f o r

as X

tends t o i n f i f i i t y i n t h e s e c t o r

-4, mt2 bO(arg~<2n-&-

(29.6)

,

bo

and

11

2

1+W

(29.7;

092

(a,A) =exp[K(l-w

m)h

(l+o(l))1

-1-2.

L++1.

t exp[K(l-w

uniforn;ly f o r ( 2 9 . 5 ) a s

tends t o i n f i n i t y i n the sector

larg A - ( 2 n - b ) l 5 b o

( 2 9 .S)

Here,

1

b0

m)12 m(l+O(l))]

.

i s an a r b i t r a r y small p o s i t i v e number,

l a r g e p o s i t i v e number, and

Ro

i s an a r b i t r a r y

LARGE EIGENVALUES

SCD

(29.9)

K'J

-l

s n [ t m + l ) ' -t2 ] d t > 0

131

.

0

In order t o d r i v e (29.4

and ( 2 9 . 7 ) , we use t h e i d e n t i t y

(29 .lo)

(cf.

( 2 6 . 3 4 ) , (26.35) and Lemma 26.1 .)

The asymptotic formula ( 2 9 . 4 )

implies t h a t l a r g e eigenvalues of Problem (28.1) do not l i e i n s e c t o r (29.6)

.

By u t i l i z i n g the asymptotic formula (29.7) i n s e c t o r ( 2 9 . 8 ) , we s h a l l

To do this, l e t us f i n d r o o t s o f t h e equation

l o c a t e l a r g e eigenvalues.

=O

We s h a l l prove the following theorem.

i n sector (29.8).

THEOFEM 29.1:

If

M i s a s u f f i c i e n t l y l a r g e p o s i t i v e number, t h e einen-

values o f problem (28.1) which a r e i n t h e domain

where p

(A1

2M

takes a l l s u f f i c i e n t l y l a r g e p o s i t i v e i n t e g e r s ,

l i m v =O p+im p

(29.13)

.

Furthermore, these einenvalues a r e simple. Proof:

Equation (29.11) i s equivalent t o

or

11

( 2 9 -14)

where

q

3( 2 i K ) s i n ( $ r ) h 2m ( lo+ (1)) =(2q+l)in

i s an a r b i t r a r y i n t e g e r .

Observe t h a t

,

a r e d v e n by

132

BOUNDARY VALUE PROBLEM

A

i n s e c t o r (29. 8 ) .

If w e s e t

11 2m

-I--

(25.15 1

p=-k

,

p=-q,

equation (29 -14) becomes 2p-1) n ) L [ ~ + F ( ~ L )( ] = 2 ’ 2K s i n ( 2 )

(29.16) where

i s holomorphic i n t h e s e c t o r

F(b)

l*g

(29.17)

f o r large

IpL(

4 26;

, and E m F(p) = 0

as

tends t o i n f i n i t y i n s e c t o r ( 2 9 . 1 7 ) , and

p

p

i s a positive integer.

aYy v i r t u e o f Theorem 3 . 4 , equation (29.16) h a s a t most one r o o t i n t h e

sec’or

(69.18)

l a w PI

< 6;

56;

9

ICL12Mo

9

Mo i s a s u f f i c i e n t l y l a r g e p o s i t i v e number. Let u s f i n d t h e r o o t of e q u a t i o n (29.16) i n s e c t o r ( 2 9 . 1 8 ) . Take Mo s o l a r g e

if

O < 6:

and

that (29.19)

.

i n s e c t o r (29.18) Denote by a t h e right-hand member of ( 2 9 . 1 6 ) , and 1 P assume t h a t -a i s i n s e c t o r (29.18). S et 2 P 1 (29.20) P p = S ~ P { I F ( w ) I ;l a r g 56: 1b1 2 p p ) 1 Assume t h a t p i s s o l a r g e t h a t p 5- sin[b:] Assume a l s o t h a t P 2 sin g r l < l men t h e d i s c d e f i n e d by

.

PI

7

2

.

(29.21)

i s i n t h e domain i29.22) Furthermore, on t h e c i r c l e

we have

.

A NON-HOMOGENEOUS BOUNDARY VALUE PROBLEM

133

I P - apl - I PI I N P ) I 2 I P - apl - I P I Pp 2 l w - a P( ( 1 - P P ) - P p a p 1- P

--pa

>O

- p a

1-2PpPP

P P

.

Hence by u t i l i z i n g t h e Rouchhls theorem we can f i n d the r o o t o f equation Denoting this r o o t by

(29.16) i n d i s c (29.21).

p

P

, we

have

and hence

This completes t h e proof of Theorem 29.1.

30.

A uniaue s o l u t i o n of a non-homogeneous boundary value problem.

In

t h i s s e c t i o n we s h a l l consider a boundary value problem

Y ~ ~ - [ P ( x ) + ~ ~ Y ,= ~~( x€ ) 5 g, E 5 .

(30.1) Let us s e t

m

-

fo(s,a,A)g(t)d5

fZ(X,a,i)J

,

X

where t h e p a t h of i n t e g r a t i o n of t h e f i r s t i n t e g r a l i s taken i n t h e s e c t o r

g2 f o r l a r g e sector

8,

nant i n

151

, while

for large

8,

and

S2

1x1

t h e p a t h of t h e second i n t e g r a l i s taken i n the

.

, respectively,

E ( h ) g i s w e l l defined, t h a t and t h a t

Since

gE 3

, and

f o and

f 2 are subdomi-

i t i s n o t d i f f i c u l t t o prove t h a t

E ( i ) g i s an e n t i r e f u n c t i o n i n

(x,a,h)

,

E ( 1 ) g i s a s o l u t i o n o f t h e d i f f e r e n t i a l equation

p - [P(x) + A ] Y = W

.

(a,a)g(x) 092 I n this s e c t i o n , we s h a l l prove t h e following theorem.

(30.3)

30.1:

If Wo,2(a,X)#0 , t h e boundary value problem (30.1) has a

uniaue s o l u t i o n

(30.4) To prove this theorem, i t is s u f f i c i e n t t o prove t h a t

134

A BOUNDARY VALUE PROBLEM

(30.5)

E(hjgE 5

and f o r every f i x e d ( a , i ) . Let us denote by SE a s e c t o r i n x-plane which i s d e f i n e d by g€ 3

f o r every

where

2

i s a small p o s i t i v e number.

Assume t h a t

,

( a ,A

,

e

I t i s eveident t h a t

and a p o s i t i v e i n t e g e r

KN

firid twc: p o s i t i v e c o n s t a n t s

and

N

zcC80

are fixed.

.

We s h a l l

such t h a t

F$

(30.6) fcr

!30.7'

larg

XI

l2

I--

This will then prove t h a t

€

I X l q J

9

E(X)g-O

in

So

.

As t h e first, case,* we assume t h a t

(30.8)

Wo,2(a,A)

In t h i s case,

and

f0

are subdominant i n

=o

-

f2 are l i n e a r l y dependent, and hence

S o u S2

.

For each

xE

zc , l e t u s

f0

denote by

and y(x)

f2

the

ray derined by

ie

(30.9; where

y(x): < = x + s e e=arg x

(30.10)

.

(O<s
( S e e F i g . 30.1.)

,

Then

E(X ) g = f o ( x , a , X ) J f 2 ( 5 , a , X ) g ( S ) d S

C m

-+ fo!x,a,X)J f 2 ( 5 , a , a ) g ( ~ ) d ~ X

- f2(x,a,A)J

m

fo(S,a,i)g(<)dS X

i s a p a t h which s t a r t s from i n f i n i t y i n t h e s e c t o r

where

K8

-

I(

and

i$i

.

and

, and

t h e p a t h o f t h e second and t h e

(See F i g . 30.2.)

By u s i n g (30.10) we can f i n d

approaches i n f i n i t y i n t h e s e c t o r

third i n t e g r a l s i s y ( x )

So

g2

easily.

To prove Theorem 30.1, we do n o t need t o c o n s i d e r this c a s e .

A NON-HOMOGENEOUS

BOUNDARY VALUE PROSLEM

135

Fig. 30.1.

Fig. 30.2. Next, assume t h a t

(30.11) I n this case,

w092( a , h ) # O . fo

t i o n s of Chapter 5 , ( 3 0 -12)

and fo

f2 a r e l i n e a r l y independent.

If we use the nota-

can be w r i t t e n a s

fo(x,a,X) = c ( a , h ) f l ( x , a , h ) - w 1-2G( a ) f 2 (x, a ,A )

,

136

A BOUNDARY VALUE PROBLEM

and

c(a,h)

i s given by (29.10).

( C f . (26.25)’ (26.29), and (26.31).)

From (29 .lo) and ( 3 @.11) we derive

( 30 As

c(a,A)#O

x

*

tends t o i n f i n i t y i n

-

% ’

(30.15) where

(m: odd)

1

(30.16)

m

1 -=b

4

(a)

,

(m: even)

,

I

(: mf2-2h) 2 mt2-2h bh ( a ) x

$1 1

2 Em(x,a) =me2

( 3 0.1‘7 )

+

:

9

l
Note t h a t the q u a n t i t i e s

as

x

Em(x,a) and

-

Sc

tends t o i n f i n i t y i n

bl

ptl ( a )

a r e independent o f

rm = {?+2b1

(m: even)

(a)

P1 ( C f . Lemma 21.1.)

as

f2(x,a,A) = w tends t o i n f i n i t y i n

x

,

sE

c =E;m(x,a)

-

.

r m c ( a , ~ ) xm[l+

*

Let us consider a new v a r i a b l e (30.21)

,

Therefore, from (30.12)’ (30.15) and (30.18),w e derive 2v ( a ) -1-i

(30.20)

m

, where (m: odd)

( 30.19)

a

6

defined by

--1 2 O(X

) I ~ X P [ E ~ ( X I, ~ )

A NON-HOMOGENEOUS BOUNDARY V L U E PROBLEM

137

By v i r t u e o f Theorem 3 . 4 , the right-hand member of (30.21) I s univalent f o r

-

(30.22) If

Mo

XES,

1x1 2 M O > O

9

I s sufficiently large.

-

as: l a r g

( 30.23)

i n <-plane.

If

T >0

9

Consider a s e c t o r

c l q 1n -

mt2

- s 2

ICI 2 5

9

I s s u f f i c i e n t l y l a r g e , the s e c t o r

I n t h e image of domain (30.22) by mapping (30.21). s u f f i c i e n t l y l a r g e , the Image of the domain

( 3 0-24)

-

xE83s

9

1x1 2%

by mapping (30.21) i s contained i n

Be

. 1

as

x

tends t o I n f i n i t y I n

zc .

I n c-plane, l e t us consider a s e c t o r

where

Note t h a t

08

i s contained

Also, if M 2 > 0 I s

A BOUNDARY VALUE PROBLEM

138

--CFT2 m+2

-T Fig. 30.3. To prove Theorem 30.1, w e s h a l l use the following lemma.

-

JJEliA 30.1: L e t p be a aiven p o s i t i v e constant. sufficiently large,

c2

are

positive

Assume t h a t Lemma 30.1 has been proved.

the unique point

such t h a t

M1=Em(xo,a)

(30.31) W e write

xo

E(A)g

in t h e form

,

-

xoE 8 ,

m, if q>0

constants independent o f

In x-plane, we s h a l l consider

I xol 2Mo

.

A NON-HOMOCENEOUS BOUNDARY VALUE PROBLEM

139

( 30.32)

Denote by I n order t o estimate the second and t h i r d i n t e g r a l s i n the right-hand member of (30.32) f o r

x€g

mate of form (30.6) f o r

, we

x€g

M2> 0 i s s u f f i c i e n t l y l a r g e .

.

use Lemma

30.1.

Then we obtain an e s t i -

Notice t h a t domain (30.24) i s i n

This proves t h a t

E(A)g 0 i n

s p l t t t i n g t h e i n t e g r a l d i f f e r e n t l y we can a l s o prove that 82

, and

hence

E(X)gE 5

Proof of Lamma 30.1: u t i l i z i n g (30.27).

By

E(1)gSO i n

The first estimate of (30.30) can be derived by In f a c t ,

=[ssm I c t s eie

7

Y(7)

cos Bl)ds]le-cI

ll-pexp(-s

0

r,(c)

=J

0

(30.36)

.

if

.

Itl-’]e -t l l d t l

(30.34)

8,

s

T+Te

ie,

P

exp[(s-

y t se

i92

T)COS

C12]ds

.

UO

A BOUNDARY VALUE PROBUN

and

This completes the proof of Lemma 30.1.

W e s h a l l denote, h e r e a f t e r , t h e right-hand member of ( 3 0 . 4 ) by R ( h ) g ; t h a t is,

(30.38)

31. Estimate o f R ( h ) g : P a r t I . R(A)g

f o r large

1x1

.

I n t h i s s e c t i o n w e s h a l l estimate

The following formula f o r

i s t h e basis of

E(h)g

our estimates: 2

t-

2

2

f2(x+ w s,a,A)g(x+ w s)ds

E(X)g = w fo(x,a,A)[ 0

i31.2)

i

2

-

f 2 ( x + w s,a,A) =urn(w 2x+ s,w

= um(w -2 x + s,w -2al,

-2 al,

...#-2b-Jlarn-l,w

...,w-2

m-1 ,O? am+\ )

-2mam+w2-”X)

.

From Theorem 7.3 we d e r i v e

where ( x + s ) m + a l ( x ts)m-lt

(31- 4 )

...+a m - , ( x + s ) t a m

= 9m + u19m-1 + . . . + ~ - l s + u m and Km(x,a)=

(31.5)

c'

, ,

(m: odd)

(m: even)

exp[-Em(x,a)]

.

Let us u t i l i z e Theorem 19.1 t o derive

where (i)

F ( s , x ) = s m + u pm-1 lull

h

(iii)

+ ...+

,


s

(iv)

;

O<s<+=

t o i n f i n i t y i n t h e sector: l a g X I -tends F1(s,h) tends t o zero uniformly f o r

lull + .**+IumI< R o as

x

tends t o zero uniformly f o r

(ii) ;,(s,X)

as

+ ...+um,ls+umt

tends t o Ro

and

9

l a g XI

I n - po ;

2.-

are a r b i t r a r y p o s i t i v e numbers, but

number depending on Ro

and

po

.

I n t h e same manner, we can derive -2 -2(m-1) Um(S,W ul,. .,w %-1 ,w-%

.

= C,(W A

-2 ul,

...,u-%

m

Mo

i s a positive

9

-2

m-1 uls

+ ...+w

m +w2-?)

--

1 1 +u2-mx)[lt~l(g,x)]^P(s,A)4expk$(o,b?du],

m

0

where (i) i ( s , x ) = s t w

1x1 2Mo

;

SCD

po

9

Po

m-1

A BOUNDARY VALUE PROBLPl

(ii) %,(s,X)

tends t o zero uniformly f o r

lull t as

X

,

O<s<+

tends t o

-b

Assume t h a t

...+ 1uml < - R ~,

~ a r g [ w ' - Y ~5.~

Po

, 1 x 1 2~~

.

gE3

1x1

(31- 8 )

;

tends t o zero uniformly f o r

lull + s

,

l a r g [ ~ ~ - ~5.l ] ( p,

tends t o i n f i n i t y i n t h e sector:

(iii) tl(s,A)

as

...+ \ u r n \< R o

, and

that

+ [all t ...t

and

(31.9)

l u g A1 I n - P o

ro and

where

~ a r g [ w ~ - Y5.~ t Po

9

are a r b i t r a r y but f i x e d p o s i t i v e numbers.

p,

(31.1), ( 3 1 . 2 ) , (31.3), (31.6) and (31.7), (31.10)

, Then from

we derive

E(A)g

1

where

= Km(x,a)Km(w -2 x,C 2 (a))Cm(ul, A

...,um+X)Cm(w-2 ul,. . . , ~ ? % t w ~ - % )

.

A

Mo such i s i n t h e domain defined by (31.8) and

From (31.10) w e conclude t h a t t h e r e exists a p o s i t i v e number t h a t , if

l a ( 2Mo

and

(x,a,i)

(31.91, we have ( W g I IC,/N,(~,~J)\

(31.12) where

C,

la1

--1 2

is a p o s i t i v e number depending only on g , ro and

Now we shall prove t h e following theorem.

po

.

ESTIMATE OF R(A)g

where

ro

143

are a r b i t r a r y but f i x e d p o s i t i v e numbers.

po

Mo g&

e x i s t p o s i t i v e numbers

C1 such t h a t ,

if

1A1

Then t h e r e

2Mo , we

have

--2 1

(31.13) Proof:

I R ( M

IC,lAl

*

I n o r d e r t o prove t h i s theorem, we must estimate

=I

...,w-aa -2 -2 x,w al, ...

urn(w-'x,

...,am+A

b;(x,al,.

we

In f a c t ,

shall use Theorem 7.3 and Theorem 19.1.

I,,,(x , al,

*.

-

w $;(w

..,a +A) m

w-'a

1'

+w2-mX) m +w2-IR) m

uniformly f o r (31.8) a s A tends t o i n f i n i t y i n s e c t o r (31.91, where Nm, and $ a r e the same a s i n formulas (31.11), (31.6) and (31.7), respectively.

Notice t h a t

P(0,A) = \ [ l + o ( l ) ]

,

F(0,A) = w 2 ? [ l + o ( l ) ]

3m-2)

( 3 1 -14)

W

092

( a J ) = N m (x,a,l)2w

[1+0(1)1

.

Hence

-

Therefore, by v i r t u e of (31.12), we can complete the proof of Theorem 31.1. Sector (31.9) i s given by (31.9')

- n + m-2 2 ~ +p o I a r g \ I n - po

.

I n Section 26 this s e c t o r was denoted by To (26.11) and (26.12).) (See Fig. 31.1.)

.

(Cf.

A BOUNDARY VALUE PROBUN

A- plane

Flg. 31.1. 32.

Estimate o f

estimates of

R ( h ) g : P a r t 11.

R(A)g

(32.1) W e remark t h a t

I n t h i s s e c t i o n , we s h a l l derive some

outside the sector

m-2 - n t 2 r t posarg h i n - p o mt-2

.

R(X)g I s meromorphic i n A-plane, and t h a t almost a l l o f

i t s poles a r e located i n a s m a l l sector (32 - 2 )

1-g

A - (2n-4n)) mt2

56,

.

( C f . Theorem 29.1.)

The d i r e c t i o n (32.3)

arg ~ = 2 n - &

is the b i s e c t o r of t h e e x t e r i o r of s e c t o r (32.1).

N g . 32.1.

(See F i g . 32.1.)

W e s h a l l first prove the following lemma.

UMMAlAl:

Assume t h a t

.

(al,. .,a m

i s fixed.

m, i n the

sector

rr-p,(argh(_nt21T-+po, m-2 mi-2

(32.4)

there exists a family of a r c s (32.5)

such t h a t (a) the h q ( t ) tions

a r e continuous functions of

t

which s a t i s f y t h e condi-

-

(32.6)

(b) no a r c of family (32.5) i n t e r s e c t s i t s e l f ; ( c ) on each a r c of familx (32.5) we have

1 1

d-

(32.7)

where C

y

a r e 'Dositive constants indewndent o f

Fig. 32.2.)

Mg. 32.2.

t

q

.

(See

146

A BOUNDARY VALUE PROBLEM

Ey v i r t u e of Theorem 3 . 4 , w e can f i n d a s u f f i c i e n t l y l a r g e p o s i t i v e number

M,

such t h a t

( 3 2 .lo)

H(h)

i s univalent i n t h e s e c t o r

larg X - ( 2 n - h ) I

Lad,< go ,

1x1 2 M o

.

The image of s e c t o r (32.10) by t h e mapping

6 =H(X)

(32.11)

contains

R

ray

if

( so 5 s < -tQ)

6 =-is

(32.12)

is sufficiently large.

so>O

Furthermore, t h e r e e x i s t s a family of

wcs

k =P

32.13)

q

(0(_7<1),

(7)

q=1,2,

...

such t h a t ( a ! ) the tions

y q ( ~ ) are continuous functions of

T

which s a t i s f y t h e condi-

(05711) 9 uniformly f o r

I =g[wq(

1

7)

1 - (2n - &)

01~51 ,

1 < 6;

0 ) J = 2 n - 5 - 6;

,

arg[ cLq 1 ) 3 = 2 n - 5 + 6 ; ;

(bj)

no a r c of family

(cl)

w e have

32.13) i n t e r s e c t s i t s e l f ;

(0<7<1)

,

PRINCIPAL PARTS OF R(k)g

(01~11) ,

M H ( I L ~ ( T ) )1J = (No+q)n -~where

No

147

is a s u f f i c i e n t l y l a r g e p o s i t i v e i n t e g e r independent of

T

and

q .

On a r c s (32.13), t h e function exp[ H( A ) ] values, s i n c e

takes purely imaginary

Theref o r e Iexp[H(X)]+lI > 1

on a r c s (32.13).

Hence, from ( 3 2 . 8 ' ) we d e r i v e (32.7) on a r c s (32.13).

Outside s e c t o r (32.10), we can use s u i t a b l e c i r c u l a r a r c s t o complete t h e proof of Lemma 32.1. I n s e c t o r (32.4) we can e s t a b l i s h an estimate

11

3IE(X)gl I C l e x p h 1 l 4 I 2 m1

( 3 2 -14) IXI

f o r large

,

where

C1 and y1

.

g, x, a, and

only on

a r e p o s i t i v e constants which depend

I n f a c t , we can derive (32.14) from ( 3 1 . 1 ) , po by u t i l i z i n g Theorems 19.1 and 20.1. Thus w e proved the

(31.2) and (31.3) following theorem.

Assume t h a t

(x,al,

...,am )

i s f i x e d , and t h a t

be an a r b i t r a r y (but f i x e d ) s m a l l p o s i t i v e constant.

po

gE 5

.

Then i n s e c t o r

(32.4), t h e r e e x i s t s a family of a r c s (32.5) such t h a t (a)

the A 9( t )

a r e continuous i n

t

and s a t i s f y conditions (32.6);

(b)

no arc of family (32.5) i n t e r s e c t s i t s e l f ;

(c)

on each a r c of family (32.5)

11

3IR(h)g(x)l I C 2 exp[Y21h12

(32.15) where

and

C2

Po

y2

a r e p o s i t i v e constants which depend only on

R(X)g

f i n d t h e p r i n c i p a l p a r t s of zeros of

W (a,X) 092 W e s h a l l assume t h a t

, and

g , x, a

*

3 3 . P r i n c i p a l p a r t s of

h

9

let

.

a t i t s poles.

I n this s e c t i o n , w e s h a l l

R(X)g a t i t s poles.

Poles of

R(h)g a r e

Hence they are the eigenvalues of problem (28.1) (al,

...,a m )

is fixed.

Let

p

.

be a f i x e d value of

A BOUNDARY VALUE PROBLEM

be t h e T a y l o r ' s expansions of

Since

f o , fl

and

f2

f o , fl and

a r e subdominant i n

f2

at

so, S1

h =p

and

respectively. S2

, respectively,

we have

( h = 0,1,2, if

p

i s a zero of

w092 ( a , a )

of m d t i p u c i t y

n-1

(33.6) where

f(x,a,p;i)

p

is a zero of

W

=

x

h=o

(a,a)

0,2

f

o,h

...,n-1) n

.

L e t u s set

h (x,a,y)(A -IA) ,

of m u l t i p l i c i t y

n

.

Then

A C0MPLE”ESS PROPERTY

(33.7)

E 3;

f ( .,a,p;A)

149

,

and

E( X ) g = -W

(33.8)

C

.

f(5,a,p;X)g(S)d5+ O ( l h - ~ ( ~ ,)

C starts from i n f i n i t y i n 8, , and approaches i n f i n i t y i n Formula (33.8) y i e l d s t h e p r i n c i p a l p a r t of R(h)g

where the path g2

2G( a ) -1 f(x,a,p;A)J

(See Flg. 30.2.)

a t a zero

p

W

of

(a,h)

of m u l t i p l i c i t y n

092

.

I n p a r t i c u l a r , the

following theorem has been proved.

Assume t h a t

:1.3 -

i s a zero of

p

W ( a , h ) of m u l t i p l i c i t y n 092 s a t i s f i e s the conditions

~

Then p i s n o t a pole of (33.9)

R(h)g

g

S,fo,h(s,a,e)g(5)d5=0

where C i n a2 .

9

h=0,1,

is a path s t a r t i n g from i n f i n i t y i n 8,

The functions

Hence, i f

p

f

(x,a,p)

0 *h

J

fo(5,a,P)g(5)ds

C

Notice t h a t

9

and approachinn i n f i n i t y

a r e given by

i s a simple r o o t of

(33-11)

...,n-1

.

W

092

(a,h)

, condition

(33.9) becomes

=o

i s an eigenfunction corresponding t o t h e eigenvalue

fo(x,a,p)

34. A comDleteness property of eiaenfunctions. I n t h i s s e c t i o n , w e s h a l l prove t h a t , i f gE 5 and condition (33.9) i s s a t i s f i e d l by g f o r every zero

p

of

W

092

, then

(a,X)

g(x)

f

0

.

We s h a l l prove the following theorem. Sumose t h a t

of

gE3;

...,am)

. m, I;f

R(X)g i s an e n t i r e function

X f o r a fixed (al, and f o r every x must be i d e n t i c a l l y eaual t o zero.

Proof:

(34.1)

By v i r t u e of Theorem 31.1, we g e t --1

, the

function

g(x)

A BOUNDARY VllLUE PROBLEM

150 for

(34.2) where

-n+ 2

po

m- 2

~ po< + arg A I n - Po

g , x, a

and

.

po

Po

9

Mo, C1 a r e p o s i t i v e con-

i s a small p o s i t i v e c o n s t a n t , and

stants depending o n l y on

ILI

9

Furthermore, by u t i l i z i n g

Theorem 32.1, we g e t

1 1

3-

(34.3)

I R ( M SC,

exp[y21AI2

m1 C2 and y2 are p o s i t i v e

on t h e family o f a r c s d e f i n e d by (32.5), where g , x, a

numbers depending o n l y on

and

po

.

Let us apply t h e Phragm6n-LindelEf Theorem t o

(34.4)

pOIarg hln+ 2

n-

~ P,

1112%

+

9

satisfies ( 3 4 . 1 ) f o r ( 3 4 . 2 ) .

R(X)g

Lle have shown t h a t

m- 2

5

choose p o s i t i v e c o n s t a n t s

C3

and

i n the sector

R(h)g

Hence, i f w e

i n a s u i t a b l e way, we g e t

--1 (34.5) on t h e boundary of ( 3 4 . 4 ) . g, x, a

and

.

po

(34.6)

m-2< C< m

and consider

-1

( 34 -10)

i n ( 3 4 . 4 ) , where

c

depend o n l y on

3

so t h a t

1 m+2 -4-) m-2

Set

3

i s an a r b i t r a r y p o s i t i v e number.

Observe t h a t

Hence,

1 m+2 2 m-2 i a -c(-) (expbe A 6,

1 mt2

11 5 e x p t - c q

is a positive constant.

have

( 34 -11)

C

and

c

be s u f f i c i e n t l y s m a l l .

F(X) = [R(h)g]A2exp[c ei a 12

i n s e c t o r ( 3 4 . 4 ) , where

i n sector (34.4).

9

n

po> 0

where we must assume t h a t

(34.8)

The two c o n s t a n t s

Choose a p o s i t i v e c o n s t a n t

I F(h)l l C 3

4 Since

1

9

1 mt2 > pC(m-2)

$+$ , we

151

A COMPLETENESS PROPERTY

.

on (34.5) for l a r g e

q

boundary of (34.4).

Hence (34.11) i s v a l i d i n domain (34.4) f o r an a r b i -

t r a r y p o s i t i v e number

Also, (34.5) and (34.10) y i e l d (34.11) on the

c ,

Letting

6

tend t o zero, we g e t

(34.12) The two estimates (34.1) and (34.12) imply t h a t

i n domain (34.4).

R(?,)g = 0

(34 - 1 3 )

as an e n t i r e f'unction of

?,

.

, we

Since (34.13) is true f o r every x

g(x)=--$R(A)g]d2

[P(x)+X]R(h)g'O

get

.

dx This completes the proof of Theorem 3.4.1. If, f o r a given

(al,. ..,am)

,

Wo,2(a,?,)

does not have any multiple

zeros, then the eigenfunctions of problem (28.1) a r e complete i n sense t h a t , i f

5 i n the

g E 3; and

J

f0(5,a,v)g(t)dS = 0

C

, then

f o r every eigenvalue

g(x)=O

.

Theorem 29.1 implies t h a t , i n

general, t h e r e exist a t most a f i n i t e number of multiple zeros of

W

(a,X)

092

.

Thus we obtain t h e theorem below.

W-

Suppose t h a t

f i n i t e number of functions

al,. gh

..,a m

in

are fixed.

Then t h e r e exist a

3 such t h a t , i f a function

g

3

s a t i s f i e s the conditions

then g

must be i d e n t i c a l l y equal t o zero.

The f b c t i o n s gh are d e r i v a t i v e s of bm(x,al,...,a w i t h respect t o

X

m-1' am + A )

a t m u l t i p l e zeros of

same as i n Theorem 33.1.

W

0,2

(a,X)

.

The path

C

i s the

CHAPTER 7

DISTRIBUTION OF ZEROS*

35.

A r e l a t i o n between zeros and eigenvalues.

(35.1) where

Let us consider

d2y/dt2- [ q ( t ) -h]y=O t

i s a real variable,

is a parameter, and

A

q(t)

i s a real-

valued function s a t i s f y i n g

Let

denote t h e s o l u t i o n of (35.1) s a t i s f y i n g t h e i n i t i a l condition

'p(t,k)

y(0) = o

(35.3) Let

N(h)

.

lim q(t)=+ t +-w

(35.2)

,

y'(0) =1

be t h e number of zeros of

i t i s known t h a t

N(X) Xl
. rp

i n the interval:

O< t < +

.

Then

has i n f i n i t e l y many p o i n t s of d i s c o n t i n u i t y

....

,

and $0

0

f o r every j

2541.) As A

.

2 rp(t,X.) d t < * J

(See, f o r example, E.A.

tends t o

h.+O J

, one

Coddington and N. Levinson [7; p .

of t h e zeros of

cp

goes t o

-4-

.

In

this chapter, w e s h a l l d e r i v e a similar result f o r subdominant s o l u t i o n s of

t h e d i f f e r e n t i a l equation (6.1)

.

Let us denote by

(354 )

Y =cp(x,al,.

.

.,a

m1

t h e unique s o l u t i o n of the d i f f e r e n t i a l equation

( 691)

*

y"

- P(x)y = 0

?his chapter i s based on Po-Fang Hsieh and

Y. Sibuya [19].

ZEROS AND EICEMTALUES

153

which s a t i s f i e s an i n i t i a l condition

where

(6.2)

m m-1 P(x)=x + a x +...+am 1

We suppose t h a t

u

(al,

...,am) .

and

a r e f i x e d complex numbers independent of

The function cp(x,a)

1.

(35.6)

.

+IPJ f o

is entire i n

.

(x,a)

We assume t h a t

*

Then

f o r every

.

(x,a)

Suppose t h a t 0

(35.8)

0

(P(xo,al,.

*

,am) = 0

*

Then (35.7) implies t h a t 0

’

(35.9)

cp (Xo 9 a1

9

...,am0 ) # o .

Therefore, we can f i n d a neighborhood

...,

z(al, am ) such t h a t ( i ) z(al,. .,am) i s holomorphic i n (ii)

.

0

z(al,

z(al,

V ;

0 ;

,...,am),al ,...,am ) = O V

.

This is a consequence of the i m p l i c i t function

We s h a l l a l s o denote by

..., am )

the s o l u t i o n In general,

and a function

...,am ) = x

identically i n theorem.

... , am ) 0

(a:,

0

(iii) cp(z(a,

of

of

V

in

( al,

“a1,.

...,am )-space.

..,am)

We c a l l

the a n a l y t i c continuation z( al,

...,%)

a zero of

. The s o l u t i o n cp(x,a) may have more than one zero. “(al, ...,a ) may be multiple-valued i n (al, ...,a )-space. m m cp(x,a)

Our main concern i n t h i s chapter i s t o c h a r a c t e r i z e subdominant s o l u t i o n s of (6.1) i n terms of s i n g u l a r i t i e s of zeros of cp(x,a) with r e s p e c t t o (al,.

..,am) .

..,am) may be multiple-valued, we s h a l l define a singul a r i t y of z(al,. ..,am) i n the following way: A zero “(al, ...,%) has a s i n d a r i t y a t ( c l , ...,am) i f there Since

z(al,.

exist a continuous curve (35 .lo)

C: (a1

,...,am 1 = ( f l ( t,..., ) fm(t))

(05 t <1)

154

DISTRIBUTION OF ZEROS

and a branch

...,am )

z(al,

(1) l i m f . ( t ) = c t+l j (2)

t h e branch

such t h a t

...,m)

( j =1,

z(al,

...,am )

,

t h i s a n a l y t i c c o n t i n u a t i o n breaks down a t Suppose t h a t

z(al,

...,am )

(cl,

...,cm) .

admits a s i n g u l a r i t y a t

t h a t t h i s s i n g u l a r i t y i s d e f i n e d by a curve (35.10).

...,am ) = -

l i m z(al,

(35.11)

Otherwise, t h e r e would be a sequence such t h a t Cp

'(xl,c)

l i r n a'n) -c n-++

.

fo

as

a

a(n)=(ain),

l i m z ( a ( n ) ) =xl#= n - +Sm

and

C

Hence t h e curve

36.

(cl,

...,cm) , and

Then

tends t o

along

c

..., ( n ) ) . Then

.

C

( n = 1 , 2 , ...)

z(a)

,

rp(xl,c)=O

would n o t d e f i n e a s i n g u l a r i t y a t

Note That i n t h i s argument w e must u s e t h e branch

singuiarity a t

,

can be continued a n a l y t i c a l l y a l o n g C

c

.

which d e f i n e s t h e

.

c

If we u s e t h e same n o t a t i o n s as i n S e c t i o n 7, t h e

Main theorem.

s e c t o r i n x-plane d e f i n e d by 2k larg x - n I m+2

(36.1) i s denoted by

8

k '

<Jmt2

The c l o s u r e o f

2k

(36.2)

-

(36.3)

k

The s e c t o r s

i.e.

% , where

. .., e l

(mod m+Z)

0,1,

cover x-plane completely.

,

n 5-m+2

(arg x - 7 1 ms2

%.

i s denoted by

Sk

A s w e d e f i n e d i n S e c t i o n 7, a s o l u t i o n

(6.1) i s s a i d t o be subdominant i n

%

tends t o i n f i n i t y i n any d i r e c t i o n

arg x = e

2k

le - m T I

(36.4)

,

<=-

if

y(x)

y(x)

t e n d s t o z e r o as

of

x

such t h a t

n

I n t h i s c h a p t e r w e s h a l l prove t h e following theorem.

36.l.:

If a z e r o

..,cm)

exists

k

such

z ( a l , . ..,am)

~f

...,am )

cp(x,a)

(cl,

...,cm)

of t h e s o l u t i o n

cp(x,a)

admits a

in (al,. ..,a m )-space, then i s subdominant i n % f o r some k . Conversely, i f t h a t (p(x,cl, ...,c ) i s subdominant i n \ , t h e n a m

singularity a t a p o i n t cp(x,cl,.

z(al,

admits a s i n m l a r i t y a t

(cl,.

there zero

..,cm) .

'he proof of this theorem w i l l be given i n S e c t i o n s 37 and 38. p r e p a r a t i o n for t h e proof, w e s h a l l make a f e w remarks on subdominant

As a

MAIN THEOREM

solutions.

155

W e know t h a t t h e d i f f e r e n t i a l equation (6.1) has t h e subdomi-

nant solution s

(36.5)

m)

Y=ldm,k(X,al,

which s a t i s f y t h e following conditions: (i) the

(ii)

bm,k(x,a)

the

m,k

admit asymptotic r e p r e s e n t a t i o n s *

(x,a)

ldm,k(x,a)

(36.6)

(x,a) ;

are entire i n

=lJ

--

-krm,k x rm,k [1+O(x 2)]exp[(-l)k+1Em(x,a)]

uniformly on each compact s e t i n

(al,.

..,am )-space

i t y i n any closed s u b s e c t o r of t h e open s e c t o r s

2bh

1

2

$W2)

Em ( x , a ) =-mf2 x

(36.8)

as

+ z

x

u-u 'k-1 %

tends t o in f in -

s+l' where

Z(mt2-2h) 1 7

mt2-2h

l
2n w=exp{i) mf2

(36 - 9 )

. .

( x , a ) i s subdominant i n % m,k i s subdominant i n Sk i f and o n l y i f cp(x,a)

The s o l u t i o n ld cp(x,a)

l i n e a r l y dependent. in

Hence t h e s o l u t i o n and Qm,k(x,a)

I n o t h e r words, t h e s o l u t i o n cp(x,a)

are

i s subdominant

Sk i f and o n l y i f

(36 .lo)

%A,k(o,a)

Let u s denote by later that

Fk(a)

Fk(a)

- &m,k(O,a)

=0

.

t h e left-hand member of (?6.lo), W e s h a l l show

i s not i d e n t i c a l l y equal t o zero.

Therefore, t h e r e a l

dimension of t h e a n a l y t i c s e t defined by (36.11)

Fk(al,.

.

..,am ) = 0

i s a t most 2 m - 2 Theorem 36.1 claims t h a t s i n g u l a r i t i e s of zeros of t h e s o l u t i o n cp(x,a) with r e s p e c t t o ( a l , . .,a ) a r e contained i n t h e m union of a f i n i t e number of t h e a n a l y t i c sets defined by (36.11), where t h e k are given by ( 3 6 . 3 ) . This means t h a t , even i f w e remove t h e s e

.

156

DISTRIBUTION OF ZEROS

s i n g u l a r p o i n t s from nected s e t .

( a l , . . . , a )-space, we s t i l l have an arc-wise con-

m

Therefore, zeros of t h e s o l u t i o n

cp(x,a)

..

I v e r s e n ' s p r o p e r t y as f u n c t i o n s of

have t h e so-called

.

In o t h e r words, t h e (al, . , am ) zeros can be continued a n a l y t i c a l l y from an a r b i t r a r y p o i n t t o another

a r b i t r a r y p o i n t along a curve which l i e s i n a g i v e n neighborhood of a pre(See M g . 36.1.)

s c r i b e d curve j o i n i n g t h e s e two points.*

This a n a l y t i c

c o n t i n u a t i o n may break down a t t h e end p o i n t of t h e c u r v e .

37.

Proof of Theorem 36.1:

...,cm)

(cl, v(x,a)

.

In t h i s s e c t i o n , we assume t h a t

i s a s i n g u l a r p o i n t o f a zero

z(al,

...,am )

of t h e s o l u t i o n

W e want t o show t h a t t h e r e e x i s t s

dominant i n

Sk

.

k such t h a t cp(x,c) As w e remarked i n S e c t i o n 3 5 , l i m z ( a ) =as

t o c along t h e curve t h e r e exist an i n t e g e r

.

( n =1,2,. .)

(37.1)

Part I.

C

h

.

which d e f i n e s t h e s i n g u l a r i t y a t c and a sequence a ( n ) = (a?', .,a ( n ) ) m

..

i s suba

tends

Hence

such t h a t l i m a( n ) = c n + sco

,

z(a("))

E S ~

. ,

( n =1,2,. .)

and (37.2)

*

lim z ( a ( " ) ) n + Sa,

In g e n e r a l , a f u n c t i o n

equation

F(f,a) = 0

f(al,

, where

T. Nishino [29; p . 601.

=a

F

.

...,am )

which i s i m p l i c i t l y d e f i n e d by an

i s e n t i r e , has t h e I v e r s e n t s p r o p e r t y .

See

PROOF OF THEORpi 36.1 i s subdominant i n one of the t h r e e s e c t o r s

We s h a l l prove t h a t rp(x,c)

.

and %+,

157

I n other words, we s h a l l prove t h a t

c

satisfies

one of the t h r e e equations (37.3)

Fh-l(c> = O

,

Fh(c) = O and

Fh+l(c) = O

.

L e t us d e r i v e a contradiction from the assumption t h a t

#'

Fh-

(37 -4)

Fh(c) f

9

o

9

Fh+l(c) #

'

Set

(37.5 Then (37.6

k

where

C (a)

(37.7)

i s defined by (21.4). Furthermore, -r (Cf. (21.211.) Wo(a) =2w m,l

.

Lst us set

Assumption (37.4) implies t h a t (37.11)

Y,(C)#O

>

YZ(C)#O

Hence t h e r e exists a neighborhood (37.12)

v,(a)#o

f o r every a E V

.

Denote by

;O

,

9

V

of

v,(a)#o

,

q(c)#O c

@#O

*

such t h a t

~ , ( a ) # o, k2(a)#o

Assume t h a t the closure o f

6 ,M

9

V

i s compact.

a s e c t o r i n x-plane which is defined by

( 3 7 -13)

6 i s a small p o s i t i v e number, and M i s a l a r g e p o s i t i v e number. The s e c t o r B i s contained i n ah Therefore, by u t i l i z i n g the asymp 6 ,M t o t i c representations (36.6) o f ( x , a ) , we obtain

where

.

m,k

DISTRIBUTION OF ZEROS

158

a€ V

uniformly f o r if M

as

x

tends t o i n f i n i t y i n

'6,M

This means t h a t ,

*

is sufficiently large,

(37.15)

for large

z(a(n') E

%

.

( n = 1 , 2 , . .)

n ,

l i m z( a ( n ) ) =a n++ z(a(n):! a r e contained e i t h e r i n t h e s e c t o r

Since

2h-1 l a r g x - ~ n 56 l

(37.17) f3r l a r g e

n

.

and

,

1x1 L M

,

the points

,

(See Fig. 37.1.)

0 2h-1

m+2

Fig. 37.1. W e s h a l l prove t h a t , i f

cannot contain any p o i n t s

M i s s u f f i c i e n t l y l a r g e , s e c t o r (37.16)

z(a(n))

.

S i m i l a r l y , we can a l s o prove t h a t

s e c t o r (37.17) cannot contain any p o i n t s large.

z(a("))

, if

M

i s sufficiently

This will l e a d u s t o a c o n t r a d i c t i o n .

L e t us f i n d t h e zeros of t h e s o l u t i o n

cp(x,a)

i n s e c t o r (37.16).

do t h i s , we s h a l l use the asymptotic r e p r e s e n t a t i o n s (36.6) o f

for

k=h

equatior.

and h + 1

, and

formula (37.8).

To

Qmlk(x,a)

From (37.8) w e d e r i v e an

PROOF OF THEOREM 36.1

159

(37.18) and R1

There e x i s t two p o s i t i v e numbers Ro

such t h a t

(37 -19)

, s i n c e (37.12) holds f o r ( x , a ) imply t h a t of U m,k

aEV

for

tations

uniformly f o r

a€V

as

x

.

Also, the asymptotic represen-

tends t o i n f i n i t y i n s e c t o r (37.16), where

r=r m,h+l-rm,h

(37.21)

aEV

r ” = h r + rm,h+l

’

*

Set

and

Then equation (37.18) i s equivalent t o (37.22) where

,

g(x,a) =(2nN)i+ $(a) N

is an a r b i t r a r y i n t e g e r .

Let u s set

Then from (37.20)

we derive

l i m H(x,a) = O

uniformly f o r

a€V

as

x

Now i t i s easy t o prove t h a t t h e

tends t o i n f i n i t y i n s e c t o r (37.16).

image o f s e c t o r (37.16) by the mapping (37.23)

u=g(x,a)

(for every fixed

a

in

V)

contains a s e c t o r (37

.a)

i n u-plane, where

Img

-6

~ - 17 1 5-69

1.

i s a s u f f i c i e n t l y small p o s i t i v e number,

s u f f i c i e n t l y l a r g e p o s i t i v e number, and in V

2z and

% is a

a r e independent o f

.

Denote by U(N)

the set i n u-plane which c o n s i s t s of p o i n t s

a

160

DISTRIBUTION OF ZEROS

(2nN)i+ $ ( a )

, where

(aEV)

i s a positive integer.

N

s u f f i c i e n t l y small.

UCN)

Observe t h a t , i f

i s sufficiently large.

(37.24).

N

and

l i m g ( x , a ) ==

U(N)

U(N) rp(x,a)

as

i n (37.24), i f

.

NfNI

i n s e c t o r (37.16) a r e tends t o i n f i n i t y i n

x

g ( z ( a ( n ) ) ,a‘n) )

i s i n s e c t o r (37.16) and

z(a(”))

is

V

i s contained i n s e c t o r

aEV , i t follows t h a t

s e c t o r (37.16)uniformly f o r f o r some

a r e mutually d i s j o i n t i f

U(N1)

Pathermore, s i n c e the zeros of

s o l u t i o n s o f (37.22), and

Then. i f

E U(N)

].(a

(4 )I

i s sufficiently large.

Assume t h a t t h e r e a r e i n f i n i t e l y many Then (37.2) implies t h a t t h e r e exist two positive integers

and

N

z(a(”))

i n s e c t o r (37.16).

z ( a ( n ) ) and

, and

z ( a( n ’ ) )

two

such t h a t

NI

(i) N # N ~ ;

,

(ii) g ( z ( a ( ” ) ) , a ( ” ) )E U ( N )

U(N’)

c

to

.

a=a(t) a(n’)

(OLtLl)

, where

W e assume t h a t

long

€18

.

t

,

a ( 1 ) = a( n ’ )

, where

U(N(t))

Note t h a t t h e curve a

and t h a t

rl

joins

tends t o

C

terminates a t along C

c

Iz(a(t))(

a(tl)

(ii) z ( a ( t ) ) i s (iii) N ( t l ) # N ( t 2 )

Let

r2

i p

to

Consider

.

z(a(t))

As

g ( z ( a ( t ) ) , a ( t ) ) of

.

c

, and

rl

that

z(a)

tends

Hence we can assume t h a t

is s u f f i c i e n t l y l a r g e .

a(t,)

of

To

Assumptions

( i ) and

such t h a t

;

s e c t o r (37.16) f o r

t l < t < t 2;

.

be the image of

rl

joins g ( z ( a ( t l ) ) , a ( t l ) ) t o

by

tradicts the f a c t that

u=g(z(a(t)),a(t))

g(z(a,(t)),a(t,))

dz(a(tl)),a(tl)) i s i n U(N(tl)) Since a l l U(N) a r e d i s j o i n t , I’2

37.2.)

.

i s a p o s i t i v e i n t e g e r depending

N(t)

( i i ) imply t h a t there e x i s t s a subarc

(i)

which j o i n s

C

i s the curve which defines the s i n g u l a r i t y a t

z ( a ( t ) ) i s i n s e c t o r (3’7.16), t h e p o i n t

t o i n f i n i t y as a(t)E V

C

be a subarc of t h e curve

a(0) = a ( n )

u-plane must be i n on

and

U(N)

are i n (37.24).

ro:

Let a(n)

, where

g ( z ( a ( ” ’ ) ) , a ( ” ’ ) )E U ( N 1 )

.

.

The curve

T,

Note t h a t

and g ( z ( a ( t , ) ) , a ( t , ) ) is i n U(N(t2)). must go o u t s i d e of U ( N ) mis con-

g ( z ( a ( t ) ) , a ( t ) )E U ( N ( t ) )

Hence s e c t o r (37.16) cannot contain any ficiently large.

.

(t,
if

.

(See Fig.

M

i s suf-

PROOF OF T H E O M 36.1

161

a- space 38.

hoof of Beorem 36.1:

.

i s subdominant i n

cp(x,c)

point of a zero

...,

Fk(a,, a,) case k = O

tion

Part 11.

.

In this section, we aesume t h a t

We want t o show t h a t

.

z ( a ) of the solution cp(x,a)

i s a singular

c

We s h a l l f i r e t show t h a t

is not i d e n t i c a l l y equal t o zero.

We s h a l l consider the

The general case can be treated i n the same way. The funcFo(a) i s defined by the left-hand member of (36.10) with k = O

.

Hence

(38.1)

=abA(o,a)

- Nm(O,a)

.

By v i r t u e of Theorem 7 . 3 ,

(38.2)

b,(x+ s,al, ...,am) =$(x,al,.

where

=

(38.3)

i’

.*,am)4m(s,ul,. ..,%)

,

(m: odd)\

exP[-Em(x,a) 3

,

(m: even)

,

and

..+ am-l(x+

( x + slm+ al(x+ s)m-l+ ,

(38.4)

= 8m +uls m-1 Hence, i f

s) + am

.

+ ...+L$&+%

Fo(a) were i d e n t i c a l l y equal t o zero, i t would follow t h a t

(38.5) This means that

@;(x,a>

- Rm(x,a) = O

for all

(x,a)

.

DISTRIBUTION OF ZEROS

162

However,

b m ( x , a ) admlts t h e asymptotic r e p r e s e n t a t i o n ( 6 . 4 ) .

Thus,

(38.6) i s impossible. I n Section 37 we proved t h a t t h e s i n g u l a r i t i e s of zeros of t h e solution

a r e contained i n t h e union o f a n a l y t i c s e t s which a r e defined,

rp(x,a)

r e s p e c t i v e l y , by F k ( & l , * . - y am ) = o

(38.7) where the

9

a r e given by ( 3 6 . 3 ) .

k

s e t s a r e a t most

.

2m-2

The r e a l dimensions of these a n a l y t i c

Hence, a s we mentioned before, w e s t i l l have an

arc-wise connected s e t , even i f we remove these s i n g u l a r p o i n t s from

... , a m)-space.

(al,

Our assumption implies t h a t

F (c)=O h

(38.8)

.

W e can f i n d a continuous curve

(38.9) in

a=a(t)

( a17...,a

m

(O
)-space such t h a t

(i) a ( l ) = c ;

(ii) F k ( a ( t ) ) # O f o r for a l l

k

05 t
given by ( 3 6 . 3 ) .

This means t h a t t h e

not s i n g u l a r p o i n t s of zeros of t h e s o l u t i o n Since

( 3 8 .lo)

\d

m,h

and

bm,h+l

Fh+l(c!#O

a(t)

cp(x,a)

( 0 5 t <1) a r e

.

a r e l i n e a r l y independent, i t follows t h a t

.

Similarly,

(38.11)

Fh-l(c)#o

By v i r t u e of (37.8) and ( 3 7 . 9 ) , t h e r e a r e t h e following two r e p r e s e n t a t i o n s

PROOF OF THEOREM 36.1

163

Let us f i x a s e c t o r (38.15) i n x-plane, where

M i s a p o s i t i v e number.

Then, by v i r t u e of (38.14) and

t h e f a c t t h a t l,j ( x , a ) i s dominant i n %-1 and 8h+1 , t h e s o l u t i o n m,h r p ( x , a ( t ) ) does not vanish on t h e boundary of s e c t o r (38.15) f o r l - p l t < l ,

is s u f f i c i e n t l y l a r g e and p is a s u f f i c i e n t l y small p o s i t i v e numNote t h a t , i f M i s b e r , where t h e choice of p may depend on M if

M

.

s u f f i c i e n t l y l a r g e , t h e function (38.16) does n o t vanish on s e c t o r (38.15).

This i s due t o the asymptotic property

Fig. 38.1.

164

DISTRIBUTION OF ZEROS

On the other hand, t h e r e are i n f i n i t e l y many zeros of t p ( x , a ( t ) ) i n 2htl 2h-111 and a r g x =the neighborhood of the two d i r e c t i o n s a r g x = m t2 m+ 2 for l - p l t < l This i s due t o the f a c t t h a t y l ( a ( t ) ) # O , y , ( a ( t ) ) # O ,

.

pl(a( t ) )# O

,

p 2 ( a (t ) )# O

for

the same method a s i n Section 37.

05 t <1

.

These zeros can be located by

This means t h a t

has i n f i n i t e l y many zeros i n s e c t o r (38.15).

cp(x,a(t))

(1- p < t < l )

These zeros must leave s e c t o r

(38.15) a s t tends t o one. Since they cannot go across t h e boundary of s e c t o r (38.15), they must tend t o i n f i n i t y i n s e c t o r (38.15). Therefore, c

i s a s i n g u l a r point of these zeros of the s o l u t i o n

cp(x,a)

.

This

completes the proof of Theorem 36.1.

Let cpI(x,a)

and

rp2(x,a)

be two s o l u t i o n s o f (6.1) s a t i s f y i n g the

i n i t i a l conditions Y(0) =I

,

y'(0)

=o

and y(0)

respectively.

Then

=o , cp(x,a)

Y'(0) = 1 can be u r i t t e n a s

rp(x,a) = w 1 ( x , a )

The condition

i s equivalent t o

On the other hand, the condition rP,(X?C)

-

m'-

lim 2 x , c x-b-

i s equivalent t o

-a

+ b2(x,a)

.

PROOF OF THEOREM 36.1

165

Therefore, we can r e s t a t e Theorem 36.1 i n terms of the r a t i o 'pl( X ,

a)

cp,o

*

I n general, l e t

yl(x,a)

t i o n s o f (6.1).

Then we can s t a t e a theorem concerning the r a t i o

and

y2(x,a)

be two l i n e a r l y independent solu-

Yl(x,a)

y,o which i s more general than Theorem 36.1. t h e readers as an exercise.

Such a generalization is l e f t t o

CHAPTER 8 A GENERAL BOUNDARY VALUE PROBLEN AND ASSOCIATED R I E M A " SURFACES*

39.

A general boundary value problem.

m

(39.1)

y " - [x +alx

m-1

t

Let v ( x , a , s , l )

be t h e s o l u t i n

f

...+a m ] y = O

which s a t i s f i e s the i n i t i a l condition Y ' ( 0 ) =TI

.

=v(x,a,l,O)

,

(p2(x,a) = d x , a , o , 1 )

Then cp(x,a,<,l) =gcpl(x,a) t Qp2(x,a)

.

If w e use the two l i n e a r l y indepen-

Y(0) = s

(39.2)

9

In p a r t i c u l a r , s e t

(39.3)

cP+,a)

dent s o l u t i o n s tp2(x,a)

b m,k ( x , a )

, respectively,

and

bm,k+l(x,a)

, we

.can write

1

cp,(x,a)

and

i n t h e following forms:

(39.4)

and

(39.5)

where

*

This chapter i s mostly based on R . Nevanlinna [ 2 7 ] and G . Elfving 181. 3 , pp. 289-3031 and E. Hille [16; Appendix D.3, pp. 654-6581.

See a l s o R . Nevanlinna 128; Chapter X I , 4$2 and

A GENERAL BOUNDARY VALUE PROBLEM

as

x

.

tends t o i n f i n i t y i n

x-plane completely, t h e r e a r e

167

-

-

80,...,Sm+l cover

Since the s e c t o r s

m t 2 functions

(39.8) which a r e defined by ( 3 9 . 7 ) .

The main problem of t h i s s e c t i o n a s w e l l a s

i n the remaining p a r t of t h i s chapter i s as follows:

PROBLEM 39.1:

Find necessary and s u f f i c i e n t conditions t h a t , f o r niven

m+2 points

bers

al

...,cmtl

co,

,...,am, S,,

on the Riemann sphere, there exist complex num-

TI,

5,,

such t h a t

TI2

(39.9) and -

dX,a,51,Q (39 .lo)

for

cp ( x , a 9 5,f

k=0,1,.

7,)

-t

'k

as

x-t=

in %

..,mtl .

This problem can be r e s t a t e d i n terms of the m + 2 functions (39.8) i n the following way: Flnd necessary and s u f f i c i e n t conditions t h a t , f o r niven

39.1':

mt2

points

co,

...,cdl

admits a s o l u t i o n

(al,

on the Riemann sphere, the system of eauations

that ... , a m , S l , ~ 2 , ~ l , ~ 2such )

(39.9) i s s a t i s f i e d .

Condition (39.9) means t h a t the two s o l u t i o n s cp(x,a,S1,V1)

are l i n e a r l y independent.

cp(x,a,S2,q2)

represents e i t h e r a complex number o r co,

...,c d l

A point m

.

c

and

on the Riemann sphere

We do not assume t h a t

are d i s t i n c t p o i n t s .

The following lemma gives us a necessary condition f o r the existence of a s o l u t i o n of system (39.11) which s a t i s f i e s condition (39.9).

after,

IIa

s o l u t i o n of system (39 .u)11 means a s o l u t i o n s a t i s f y i n g (39.9)

39.1:

.

In order t h a t t h e r e exist a s o l u t i o n of system (39.11), i t i s

necessary t h a t (39.12)

Here-

ck#\+l

where c H 2 = c o .

(k=O,l,...,*l)

9

168

A C m R A I , BOUNDARY VALUE PROBLEM

Proof:

Let us regard

(39 -13)

as p o i n t s on the Riemann sphere.

Then i f

c = cj+l

f o r some k

, and

if

t h e r e i s a s o l u t i o n of (39.11), we must have

(39 -14)

f o r such k

f k ( a ) =fk+l(a)

.

However, ( 3 . 1 4 ) i s impossible, s i n c e

(x,a) m,k This proves Lemma 39.1.

l i n e a r l y independent.

and

ldm,k+l(x,a)

are

The following lemma w i l l present a less obvious necessary condition. 39.2:

In order t h a t t h e r e exist a s o l u t i o n of system (39.11), i t i s

necessary t h a t the s e t

{c,,

. ..,c mtl )

contains a t l e a s t t h r e e d i s t i n c t

points.

The necessary condition (39.12) implies t h a t the s e t

Proof:

contains a t l e a s t two d i s t i n c t points.

set

{co,

...,cmtl ]

Lemma 39.1,

(39.15)

(c,,

...,c mtl ]

Assume f o r a c o n t r a d i c t i o n t h a t the

c o n s i s t s of e x a c t l y two p o i n t s .

Then by v i r t u e of

m must be even, and

[

c0=c2=c4= c1=c3=c5=

... = c & = ... = cm-2 = cm ...=

'

-

c 2 h + l - ' ' ' = c m-1 - 'm+l

*

This means t h a t

( a ) ldm,2h('x, a )

( h = 0,

1 ...,@

a r e l i n e a r l y dependent,

and (b)

bm,2h+l( x, a )

( h = 0,

Thie is impossible.

. ..,-)12

a r e l i n e a r l y dependent.

( C f . the proof of Theorem 21.4.)

Lemmas 39.1 and 39.2 give us two necessary conditions f o r the existence o f a s o l u t i o n of system (39.11).

We s h a l l show t h a t these two condi-

tions are also s u f f i c i e n t .

THEQRpi 39L: In order t h a t t h e r e exist a s o l u t i o n of system (39.11), is necessary and s u f f i c i e n t t h a t (i) % # ~ k + ~( k = O , and (ii) the s e t f c O ,

-

...,mt1;

...,cU t t l )

In the case

m =1

Theorem 39.1 means t h a t

, we

cmt2= co)

9

contains a t l e a s t t h r e e d i s t i n c t p o i n t s . have

co, c1

m + 1= 2 and

c2

, and

hence condition (ii)of

are d i s t i n c t p o i n t s on t h e

it

A GENERAL BOUNDARY V A L E PROBLEM

Riemann sphere.

169

Also, i n this case, t h e three points

, since

on the Riemann sphere a r e d i s t i n c t for every value of

a

of

Therefore, f o r each

41,k(x,a)

fixed value of

(k =0,1,2)

, there

a

T(f) =

a r e l i n e a r l y independent.

any two

e x i s t s a transformation

-

5,f v1 5 9 - 92

9

Slv,

- 529,f

0

9

such t h a t ck=T(fk(a))

(k=0,1,2)

.

This proves t h a t conditions (i)and (ii)of Theorem 39.1 are s u f f i c i e n t i f m=l

.

I n o r d e r t o i n v e s t i g a t e the case m = 2 , l e t us make t h e following observation. M r s t of a l l , by using t h e n o t a t i o n f k ( a ) defined by

(39.13), we can write (39.11) a s (39.11') Furthermore, i f we use the d e f i n i t i o n Um,k(x9a) Of

bm,k

(x,a)

(39 -16)

, we

'bm(W -k X,G k ( a ) )

obtain

fk(a) = w

-kf(Gk ( a ) )

..

(k = 0,1, .,mtl)

,

(39.17)

(39 -18)

Gk ( a)

I n the case m = 2

= ( w-kal, w-=a2,

...,"-*am

,

w = exp

L+i

1

.

,

(39.19)

where U(a,x) this case,

i s given by i n t e g r a l (8.6).

(Cf.(8.10) .)

Therefore, i n

( 39.21)

and

GENERAL BOUNDARY VALUE PROBLEM

A

170

c

3

(39 2 2 )

c0 =

o,

is a r b i t r a r y .

cl=l,

c2=m

’

Condition (i) of Theorem 39.1 r e q u i r e s t h a t

c3#0,

.

I t i s easy t o see t h a t t h e case when

co

, c1

and

a r e d i s t i n c t points 2 on the Riemann sphere can be reduced t o t h e s p e c i a l case (39.21). c

System (3.1l1) becomes, i n this case,

Prom Hence

(Eo) w e g e t T1=51fo(a)

( 5 ) ylelds 5, 5,

-= Thus we obtain from

*

,

and from

(E2) we g e t T2=5,f2(a)

f1W - f2b) flM -fob) ’ (E,)

t h e following equation:

See, f o r example, M. Abramowitz and I . A . S t e w [l; p. 687, 19-3-51.

.

A GENERAL BOUNDARY VALUE PROBLEM

I

171

r

[

Jn

Ut(b,O) =-

1 1 -b-'2 % ( a + + b )

Therefore, from (39.20) and (39.24), i t follows t h a t

(39 -25)

f(0,a2) =-+

Furthermore, i f

m=2

I

, we

r(2 + 1- a )

r(-+4 42

have

.

w = e x p [ i2- n ] = i

L

.

Hence (39.16) y i e l d s

.

1 f3(0,a2) = itan[--n(l4 a2)]f(0,a,)

I n deriving (39 2 6 1 , we have u t i l i z e d t h e well-known formulas

and s i n [ -1- n ( l + az) J = c o s [1 - n ( l - a2) J

4

4

.

I n s e r t i n g (39.26) i n t o (39.23), we obtain

I

-

}

2 1 i t a n [1 - n ( l - a2) ]

(39 2 7 )

1 l + it a n [ - n ( l - a 2 ) ]

=c3.

4

This equation has r o o t s f o r every value of c

3

=m

.

Note t h a t

i

and

c = O and 3 -iare the P i c a r d ' s exceptional values f o r c3

except f o r

.

This shows t h a t , i n t h e case m = 2 , conditions ( i )and (ii) of Theorem 39.1 a r e a l s o s u f f i c i e n t . F'urthermore, i f we regard a2 a s a tan x

c , we can show t h a t 3 p o i n t s of this function .* h c t i o n of

c =O

3

and

8

are logarithmic branch-

* me author obtained this information from Professor Tosihusa Kimura, University of Tokyo. At c =1, t h e r e a r e s i n g u l a r i t i e s o f d i f f e r e n t type. 3 See Section 42.

A GENERAL BOUNDARY VALUE PROBLEM

172

The proof of Theorem 39.1 f o r

m23

i s not simple.

We s h a l l explain

how t o prove the s u f f i c i e n c y of conditions ( i )and ( i i )of Theorem 39.1 i n

t h e following seven s e c t i o n s . 40.

In order t o f i n d a much deeper meaning

Associated uemann s u r f a c e s .

o f Problem 39.1, l e t u s set

F(x) =

(4o.U

cp( x , a

,I,

,vl)

c p b , a ,5,, T2)

and

where

a,

5

and

a r e f i x e d so t h a t

71

W e shall prove the lemma below.

m: The ( i ) F(x) (ii)

kction

Ft(x)

(v)

F(x)+%

as

F(x)

f.F9x3

i s eaual t o (40.5) Proof:

-2[x

m

i s simple;

x tends t o i n f i n i t y i n

the Schwarzian of

(40.4)

s a t i s f i e s t h e followinn conditions: over x-plane;

x

does n o t have zeros i n x-Dlane;

(iii) every pole of

(iv)

F(x)

i s meromorphic i n

% , where

k=0,1,

...,In+,

;

F(x) :

=( #j

- 51 (F*F” j2

x

+ alx m-1 + ...+ am] , h,

{F,x)=-2[x

m +alxm-1

+ ...+am] .

We have shown ( i v ) already i n Section 39.

Conditions (i)and (iii)

.

are immediate consequences of d e f i n i t i o n (40.1) of F(x) Property ( v ) of the Schwarzian of F(x) can be derived by a s t r a i g h t - f o w a r d cornputation.* Property (ii)can be proved by u t i l i z i n g the formula FI(x)

=

,T,)cp(x , a , 5,

cp I ( x ,a

The numerator i s the Wronskian of Wonskian i s independent of ‘ h i s completes the proof of

*

x

,

91,)

-d

x , a ,5,,

Tll)rd(x,a 9 5,,T2)

[cp ( x 9 a A,, 1,) l2 cp(x,a,12,v2) and qdx,a,cl,Til) This and hence this i s equal t o ‘Ills2- T25,

.

Ismma 40.1.

See, f o r example, E. Hille [16; Appendix D.1, Theorem D . 1 . 1 ,

p.

6481.

.

ASSOCIATED RIEMA" SURFACES

173

Let us denote by

(40.6)

x=G(z)

t h e i n v e r s e function of (40.7)

z=F(x)

Since G ( z )

z a s a point on t h e

i s not single-valued, we must regard

Riemann surface face R

.

R

of t h e inverse function of

.

F(x)

This Riemann sur-

i s conformally equivalent t o x-plane, and such a conformal mapping

i s given by ( 4 0 . 7 ) .

i s simply connected, and t h a t R i s of parabolic type. The Riemann surface R can be regarded as a coveri n g surface over the Riemann sphere. Properties (ii)and (iii)o f F(x) R

This means t h a t

does n o t have any algebraic branch-

imply t h a t t h e covering surface R

p o i n t s Furthemore, property ( i v ) means t h a t branch-point over each base p o i n t

k

such t h a t

and

ck

jfk

, the

\

.

R

Even i f

admits a logarithmic

cj=ck

f o r some

logarithmic branch-points corresponding t o

a r e d i s t i n c t on t h e covering surface

branch-points l i e over the same base p o i n t

R

cj =\

these r e s u l t s i n the following lemma.

.

and

j

c

j

These two logarithmic

.

We s h a l l summarize

LO.* The coverinn surface R which i s defined by the i n v e r s e funct i o n o f F(x) s a t i s f i e s the conditions below: (i) R i s conformdly equivalent t o x-plane; (ii) R does not have any s i n n u l a r points o t h e r than the loaarithmic branch-points over the base points co,cl, cm t l

...,

Remark 1: Property (ii)of the surface R

follows from t h e f a c t t h a t the

transcendental s i n g u l a r i t i e s of the inverse of a meromorphic function a r e t h e asymptotic value8 of t h e meromorphic function a s

x

tends t o i n f i n i t y .

The following lemma shows t h a t F(x) does not have t h e P i c a r d ' s exceptional values. Note t h a t the P i c a r d ' s exceptional values of a mero-

Remark 2:

morphic h c t i o n must belong t o t h e s e t o f i t s asymptotic values as x tends t o i n f i n i t y . A point

(40.8)

xo

of x-plane

F(x) = \

i f and only i f

(40.9)

b,&(Xo,a)

=o

*

s a t i s f i e s the eauation

A GENE2U.L BOUNDARY VATXI3 PROBLEM

174

Proof:

If we use t h e n o t a t i o n s given by (39.31, (39.4) and (39.51, w e can

write

CP(X,~,S,TI)

as

(40.10)

where

(40.11) and

(40.12)

Observe t h a t

(40.14) and

( 40.15 ) Note t h a t t h e two s o l u t i o n s cp(x,a,S1,T1) independent.

, there

q(x,a,s2,l2)

are linearly

A straight-forward computation proves Lemma 40.3.

‘fie Stokes m u l t i p l i e r s a

and

exists an i n t e g e r

%(a) k

have t h e property t h a t , f o r each fixed

such t h a t

%(a) # O

.

( C f . Theorem 21.4.)

Therefore, by the same method as i n Section 37, we can prove t h a t bm(x,a)

has i n f i n i t e l y many zeros for each f i x e d

not have t h e P i c a r d ’ s exceptional values. below.

LO.&

a

.

This proves t h a t F ( x ) does Thus, we have proved t h e lemma

F(x) does n o t have t h e P i c a r d ’ s exceptional v a l u e s .

‘The main i d e a of t h e proof of Theorem 39.1 i s t o e s t a b l i s h t h e converse of Lemma 40.2. R. Nevanlinna [ 271 considered t h e following problem.

yI,. . . , y

PROBLEM LO.1: sphere, where

922

,

a coverinn s u r f a c e R (i) R and -

be q and l e t nl,

q

d i s t i n c t p o i n t s on t h e Riemann

...,nq

be p o s i t i v e i n t e p e r s .

of t h e Riemann sphere such t h a t

is s i m l ~connected,

Construct

175

ASSOCIATED RIEMANN SURFACES

admits no sinmilar points other than e x a c t l y n

(ii) R

j

logarithmic

.

where j =1,...,q j'Assume t h a t t h e r e exists such a covering surface R f o r a given d a t a

branch-points over t h e base p o i n t y

(yl,

...,yq;

.

Then Nevanlinna a l s o proved t h e theorem below.* nl,. ..,n ) q The coverinp: surface 62 i s conformally equivalent t o the Furthemore, &t g(x)

complex plane.

be a meromomhic function such t h a t

z =g(x)

(40.16)

i s a conformal mapping from x-plane onto R

of

g(x)

.

Then the Schwarzian

i s a pobnomial i n x whose degree i s q p = 2 n. j=1

(40.17)

p- 2

{g,x]

, where

.

Theorem 40.1 explains the r e l a t i o n between Problems 39.1 and 40.1.

i s known that**

It

g ( x ) i s represented by a r a t i o of two l i n e a r l y indepen-

dent s o l u t i o n s o f 1

y" t z{g,x}y = 0

(40.18)

{g,x} i s a polynomial i n

Since

. x

of degree p - 2

, equation

(40.18) can

be reduced t o

(40.19)

i f w e replace

by

x

cx

, where

c,al,. ..,a

numbers. I n the case p = 2

, we

q=2,

P-2

are s u i t a b l e complex

must have

n =n =1. 1 2

Furthermore, by using t h e d i f f e r e n t i a l equation

yll-y=O

, we

can conclude

that (40.21) where

Lt

and

i

g ( x ) -IL+

as

Re[x]-*+

g ( x ) +L-

as

Re[x]+-=

-

L

covering surface R

, ,

a r e c e r t a i n points on t h e Riemann sphere.

Since the

s a t i s f i e s condition (ii)of Problem 40.1, i t follows

that

*

**

The main i d e a of the proof o f Theorem 40.1 i s i l l u s t r a t e d i n Section 45.

See, f o r example, E. Hille [16; Appendix D.1, Theorem D.l.l, p. 6481.

A GENERAL BOUNDARY VALUE PROBLEN

176

(40.23)

p> 2

In the case

(40.24)

or L =y

, we

..

k = 0,1,. ,p-1,

Theref ore,

-

So,

, we

...,SPm1

L =y + L =y - 1

.

have

.

2x

or

g(x) = e

have as

and

gk, a r e

( 4 0- 2 5 )

=m

-2x

g(x)+ck

and the s e c t o r s

2

g(x) = e

either

{

L =y

y1 = 0 and y

In p a r t i c u l a r , i f

where

{

either

(40.22)

%,

x-*m i n co,

..., cP-1

are p o i n t s on t h e Riemann sphere.

t h e same a s w e defined i n Section 7 with cover x-plane completely.

ck E I y l ,

and t h e r e e x i s t , f o r each

-.,Y,I j

-

(k = 0 ,

, exactly

n

-

*

Pp-1)

.

Furthermore, 9

distinct indices

j

m=p- 2

k

such t h a t

k' - Y j ' W e s h a l l prove t h e s u f f i c i e n c y of conditions (i)and (ii)of Theorem

39.1 by constructing a s u i t a b l e covering s u r f a c e 62 over t h e Riemann sphere. W e s h a l l explain such a method by u t i l i z i n g a very simple case. The proof f o r the g e n e r a l case i s similar, and will be l e f t t o t h e readers.*

a. Construction

.

of a coverinp s u r f a c e R

In t h i s s e c t i o n and s e c t i o n s

&, 45, and 46, we s h a l l consider the case when (41J)

m=3

and

(41.2)

i

c0 =

o,

cl=l, c2=1+i, c3=i, c,+ =1+i

,

and w e s h a l l prove t h e e x i s t e n c e o f a s o l u t i o n of system (39.11). t h a t the q u a n t i t i e s

co, cl,

c2, c3

(ii)of Theorem 39.1.

*

see, for example, G. a f v i n g [ 81.

and

c

4

Note

s a t i s f y conditions (i)and

177

A COVERING SURFACE

Let us consider a pentagon whose v e r t i c e s a r e denoted by A3

and

A

4 '

{Ao,A1,%,A3,A4] (41.3) where

(See Fig. 4 1 . 1 . )

A o , A1,

A2,

from t h e s e t

=%

(k=0,1,2,3,4)

a r e given by ( 4 1 . 2 ) . In this s e c t i o n , we s h a l l construct a

ck

covering s u r f a c e 62,

over t h e Riemann sphere such t h a t

i s a simply connected and open Riemann surface;

(ii) Ro A.

c(A)

i n t o t h e Riemann sphere by c(%)

(i) Ro

Define a mapping

contains a domain which is homeomorphic t o t h e pentagon

%%SA4

i n such a way t h a t each v e r t e x

mic branch-point of

Ro

over t h e base p o i n t

%

corresponds t o a l o g a r i t h c ( %)

, and

t h a t t h e orien-

t a t i o n i s preserved; (iii) Ro admits no s i n a u l a r p o i n t s o t h e r than the f i v e loearithmic branch-points described i n (ii)

.

Flg.

41.1.

There a r e f o u r d i s t i n c t p o i n t s (41.4)

0 ,

1 , l+i,i

.

Consider t h e square on the FEemann sphere {co,c1,c2,c3,c4] whose v e r t i c e s are a t t h e f o u r p o i n t s (41.4). (See Fig. 4 1 . 2 . )

i n the set

Fig. 0.2.

A G m R A L BOUNDARY VALUE PROBLEM

178

This square d i v i d e s t h e Riemann sphere i n t o two domains D1 assume t h a t t h e p o i n t a t i n f i n i t y belongs t o

D2

.

triangle

AoA3A4

'I"nis s u b d i v i s i o n y i e l d s a t r a p e z o i d

.

The f o u r v e r t i c e s

Ao,

A3

A,,

A3

and

,

3'

of t h e t r a p e -

Similarly

of t h e t r i a n g l e a r e l o c a t e d on t h e

A

4

boundary i n t h e counter-clockwise s e n s e .

, c(%)

A

and a

zoid a r e l o c a t e d on t h e boundary i n t h e counter-clockwise s e n s e . the three vertices

to

A,

AoA1A2A3

and

A1,

We

(See F i g . 41.2.)

Let us subdivide t h e pentagon by t h e line-segment j o i n i n g (See n g . 41.1.)

.

D2

and

On t h e o t h e r hand, t h e f o u r

,

c(A ) = i a r e t h e ver3 t i c e s of t h e square on t h e Riemann s p h e r e which i s given by F i g . 41.2.

points

c(A1) = O

=1

c(AZ) = l + i and

These f o u r p o i n t s are l o c a t e d on t h e boundary o f clockwise s e n s e . c ( k ) = 1t i

4

Similarly, the three points

a r e l o c a t e d on t h e boundary of

D1 i n t h e counter-

c(Ao) = O

,

c(A3) = i

, and

i n t h e counter-clockwise

D2

This o b s e r v a t i o n shows t h a t mapping (41.3) can be extended t o a

senee.

homeomorphism from t h e pentagon onto t h e Riemann sphere c u t a l o n g t h e a r c c o n s i s t i n g o f t h r e e line-segments 1

,

lfi

SO

and

lti

to

i

L1

, L2

L3

which j o i n

Since

LyLyL3

and

, respectively.

0

to

1

,

i s a c u t on

t h e Riernann sphere, each of t h e s e t h r e e line-segments r e p r e s e n t s two l i n e segments. where

D2

i

For example,

1 9 1

.

L1

r e p r e s e n t s two line-segments

, and , L2,2 , and

i s on t h e boundary o f

Similarly, we define

41.3.)

.

L2,1

D1

L

192 L 311

L 191 and L 1 , 2 ' i s on t h e boundary o f

'

L3,2

.

(See F i g .

l+i

2 . 3 '

I.

4.2

1

F i g . 41.3. Let u s denote by the vertex

A. to J The homeomorphism from t h e pentagon onto t h e Riemann

4, .

s p h e r e , c u t along

(41.5)

L"L'L

Aj-%

t h e line-segment j o i n i n g t h e v e r t e x

, maps

t h e l i n e segments

1 2 3 - - $'L1 ' 4%' A2A3 ' A3A4'

and

ALAo

A COVERING SURFACE

179

e i t h e r onto

L1,l’ L2,1

(41.6)

’

U

L3,1

L3,2

9

9

a d

L2 , 2

’

L1,2

r e s p e c t i v e l y , o r onto

(41.7)

’

5’2

’ L3,2 ’

L2,2

L3,1

’ and

’

L2,1‘%,1

In order t o preserve t h e o r i e n t a t i o n , we must choose ( i J . 6 )

respectively.

a s t h e image of the f i v e line-segments (41.5).

The end-points of these

five arcs ( u . 6 ) are

(41 -8)

0

The p o i n t

D1

, 1 , lti , i , and

l+i appears twice.

, while

l+i

.

The first p o i n t

lii i s on t h e boundary of l+i i s on the o t h e r s i d e of t h e c u t . Thus

t h e second p o i n t

these two p o i n t s w i l l eventually represent two d i s t i n c t p o i n t s over t h e

same base p o i n t

l+i

.

The next s t e p i s t o a t t a c h a s u i t a b l e domain along each of the f i v e a r c s (41.6) so t h a t t h e f i v e p o i n t s (iJ.8) points.

become logarithmic branch-

For example,

Such s u i t a b l e domains a r e c a l l e d logarithmic-ends.

consider t h e a r c

.

L2,yL1,2

I n order t o c o n s t r u c t a logarithmic-end

along this a r c , we s h a l l u s e countably many copies of the Riemann sphere which is c u t along

.

L2v5

We denote by

S

( j =1,2,.

j of t h e Riemann sphere each of which i s c u t along an a r c

is a copy of

LyL1

which a r e copies of two a r c s

‘j ’2

4

j

and

.

S

The surface

, where

4,

j ’2 i s on t h e boundary of

D

j

j’2

4,

j

‘

these copies

.

Lj has two domains D

, respectively.

D1 and D2

..) j

The c u t

C

j i s on t h e boundary of

(See Fig. 41.4.)

The a r c and

D

&

j

j,2

represents D

j

.

, and

ir-----

0

I , Fig. 41.4.

S1 t o L2,yL1,2 by i d e n t i f y i n g C 191 with L z , y a l s o a t t a c h Sj+l t o S by i d e n t i f y i n g Cj+l,l with &tJ,2 9

W e attach

j

4,2. We

A GENERAL BOUNDARY VALUE PROBLESI

180

where

j =1,2,

L2,2 "L 1,2

.

....

I n this manner, w e can a t t a c h a logarithmic-end along

This process i s t o p o l o g i c a l l y equivalent t o a t t a c h i n g an open

semi-disc t o t h e pentagon

-

along

Ao4%A3A4

A4Ao

.

(See Fig. 41.5.)

Fig. 41.5. I n order t o c o n s t r u c t a s u i t a b l e L1,l * w e s h a l l again consider countably many logarithmic-end along this a r c , Let us next consider t h e arc

- ....

.

We denote by 3 ( j =1,2, ) L, j these copies of the Riemann sphere, each o f which i s c u t along an a r c .t

copies o f the Riemann sphere, c u t along The a r c

-

l:

and

i s a copy of t h e a r c

J-

, where

4

j' ,1 j ,2 i s on t h e lower s i d e o f

L, I

.

,j ,1

t,

j

.

The c u t

.c",

j

r e p r e s e n t s two a r c s

J

i s on t h e upper siae of

-

j

(See Mg. a . 6 . )

, while

j ,2

N

Fig. 0 . 6 .

we j o i n

9,

to

by i d e n t i f y i n g

by i d e n t i f y i n g i to S j+l,2 'j+l j w e can a t t a c h a logarithmic-end along

Ao% .

.

d

with L,l 1,2_ and 4, ,1 , where

%,1

.

We a l s o a t t a c h

j = 1,2,

. ...

Thus

This process i s again topo-

l o g i c a l l y equivalent t o a t t a c h i n g an open semi-disc t o t h e pentagon

Ao%%A3A4

along

(See Fig. 41.5.) S i m i l a r l y , w e can c o n s t r u c t a

logarithmic-end along each o f t h r e e o t h e r a r c s of ( a . 6 ) .

Thus w e

REMARKS construct a covering surface Ro

181

which s a t i s f i e s conditions ( i ) ,(ii) and

(iii).

42.

Remarks on non-uniaueness and s i n w l a r i t i e s of s o l u t i o n s .

I n the

preceding s e c t i o n , we constructed a covering surface sphere s o t h a t

Ro

over the Riemann Ro s a t i s f i e s the t h r e e conditions ( i ) ,(ii) and (iii)

.

I n this s e c t i o n , we s h a l l present some examples which will show t h a t such covering surfaces are not unique.

Example I:

which is c u t along

(l)"L(l)u L1,l 2,l

S(l) and S ( 2 ) o f the Riemann sphere The c u t on S(') represents two a r c s

Consider two copies

eLyL3

L3,1 and

s e n t s two a r c s

.

(')"

( 2 ) WL(2) +2) 2,1 3 , 1

%,1

L ( l ) and L(2) are copies o f j ,k j ,k (See Fig. 42.1.)

, while

(')"

4,2 L2,2

L

Lg,2

( 2 ) " ~ ( ~( 2) ) ~ 4,2 2,2 L3,2

and j,k

the c u t on

'

where

i

j=1,2,3

I

.

and

S(2)

repre-

The two a r c s k=1,2

.

l+i

Fig. 42.1.

.

(2)uL(2)u (2) Thus we have constructed a covering s u r f a c e Q over 2 , l L3,1 the Riemann sphere. There e x i s t s a homeomorphism from t h e pentagon

'1,l

Ao%A2A

A

34

onto

(42.1)

such t h a t t h e images of f i v e line-segments

- - - ' 0 %

' %% ' 2'3'

' 3'4'

and

Eo 4

are respectively

(42 *2)

Ey a t t a c h i n g s u i t a b l e logarithmic-ends t o Q along t h e f i v e arcs (42.2), we can construct a covering surface 61 which s a t i s f i e s the same conditions

A GENERAL BOUNDARY VALUE PROBLEM

182

(I), (ii)and (iii)a s t h e s u r f a c e Ro

.

from Ro EkamDle A ~ .A

11:

does.

L e t us subdivide t h e pentagon

However,

Ao4A2A3A4

R

is d i f f e r e n t

by the line-segment

(See ~ F i g . 42.2.) A0

Fig. 42.2. Consider two copies

S"(l)and

of t h e Riemann sphere, where

s(2)

i s c u t along an a r c which is a copy o f

an a r c which i s a copy of

of

L

j ,k

,

respectively.

q L y L 3

.

LyL2

, while s"(2)

The c u t on

(See F i g . 42.3.)

+)

i s cut along

s"(l)r e p r e s e n t s two

REMARKS

183

There e x i s t s a homeomorphism from t h e t r i a n g l e

onto

A,%%

such

t h a t the images of the t h r e e line-segments

-

- -

(42.3)

’

AoAl

!LA2

and

%Ao

are, respectively, (42.4)

There also e x i s t s a homeomorphism from t h e trapezoid

onto

AoA2A3A4

4 2 ) S

such t h a t t h e images o f t h e f o u r line-segments

, A2A3 , A3A4

AoA2

(42.5)

and

A A

40

are, respectively, (42.6)

Let us a t t a c h -(2)“-(2) L2,1

4,l

-S( 2 )

to

by i d e n t i f y i n g

-(l) 5 .2

with 2.2

I n this way, we construct a covering s u r f a c e

B“

over the

There e x i s t s a homeomorphism from t h e pentagon

Riemann sphere.

Ao%A2A 3A4

such t h a t the images of t h e f i v e line-segments ( 4 2 . 1 ) a r e , respec-

onto tively, (42.7)

.-

By a t t a c h i n g s u i t a b l e logarithmic-ends t o

B along t h e f i v e a r c s ( 4 2 . 7 ) ’

we can c o n s t r u c t a covering s u r f a c e which s a t i s f i e s the same conditions (i), (ii)and (iii)a s t h e s u r f a c e Ro

, but

i s d i f f e r e n t from R o I n Section 39, we i n v e s t i g a t e d the cases m = l and m = 2

39.1.

I n p a r t i c u l a r , i n case

m= 2

equation (39.27)

If we regard

i

a

1- i t a n [1 - n ( l - az) ] 1

rxc3. , equations

.

o f Theorem

(39.11) were reduced t o the

1+ i tan[-n(l- a2) 1 4

the points = O and c = a a r e 3 ’ c3 3 logarithmic branch-points of this f u n c t i o n . This means t h a t equations 2

a s a function of

c

(39.11) may have i n f i n i t e l y many s o l u t i o n s f o r f i x e d values of The non-uniqueness o f s o l u t i o n s of (3.11) i s c l e a r l y

C0’C1, * .,c mtl r e l a t e d with the non-uniqueness of covering s u r f a c e s t h a t w e have shown

by Examples I and 11.

The function a2 of t h e q u a n t i t y However, the condition

c

3

is holomorphic a t

c =I 3

.

A

12%

GENERAL BOUNDARY VALUE PROBLEM

y12- T2Tllf 0

(39.9) i s violated i f

Z ( 1 - t a2) =0,-1,-2

I

,... ,

1

,... .

OI’

- ( 3 + a2) =O,-1,-2

4

Hence t h e r e a r e c e r t a i n s i n g u l a r i t i e s of s o l u t i o n s o f (39.11)a t Let us consider the case when m = 3

,

c0 = o

(42 .e;j

cl=l

,

, and

c2=l+i

,

c3=€i

,

c 3 =1

.

c4=l+i

uhere

(42.9)

.

O<€
F’cr ehch value o f

c

i n i n t e r v a l ( 4 2 . 9 ) , q u a n t i t i e s (42.8) s a t i s f y condi-

t i o n s ( i )a n d (ii)o f Theorem 39.1. Furthermore, i f

O<

€51 ,

we can

construct a covering surface by using the same method as i n Section P r e c i s e l y speaking, w e s t a r t with the pentagon

c(%) =ck

([email protected]) where

ck

AoA3 .

and t h e mapping

(k=0,1,2,3,4)

a r e given by (42.8)

line-segmen t

Ao4%A3A4

a.

.

Then w e subdivide the pentagon by the

W e extend mapping (42.10) t o a homeomorphism f r o m

the pentagon onto the R i e m a n n sphere which i s c u t along t h e t h r e e l i n e -

0

segments j o i n i n g

to

1

,

1 to

lti

, and

(See Fig. 42.4.)

l+i t o

ci

, respectively.

.

Fig. 42.4. By attaching s u i t a b l e logarithmic-ends t o this surface, w e can construct a

desired covering s u r f a c e . Note t h a t ,

at

E

=0

,

However, this process breaks down a t

c ( Ao) = c ( A3) = 0

.

c=O

.

Hence t h e subdivision of t h e

A pentagon

Ao%%A3AI,

ON THE GENERAL CASE

RplARK

by

A A

does not work.

0 3

185

This phenomenon would y i e l d

a s i n g u l a r i t y which i s s i m i l a r t o the s i n g u l a r i t i e s a t

c3

=1 o f the case

m = 2 .

43.

A remark on the g e n e r a l case.

AoqA2A3A m+2

4 '

vertices

In Section &L,

we considered a pentagon

In the general case, we consider a r e g u l a r polygon

.

Ao,4,...,~l

i n t h e counter-clockwise sense.

We assume t h a t

Ao,

Define a mapping

..,hl]

i n t o the Riemann sphere. {Ao,. s p e c i a l case of t h e problem below.

Km

with

...,bl are arranged

c(A)

from t h e s e t

The problem i n Section 41 i s a

PROBLEM L3.1 : Construct a covering surface R

over the Riemann sphere s o

that ( i ) 63

i s a simply connected and open Riemann surface; contains a domain which i s homeomorphic t o t h e polypon

(ii) R

such a way t h a t each v e r t e x

of

over t h e base p o i n t

R

(iii) R

4,

Km

corresponds t o a logarithmic branch-point

c(pk)

, and

t h a t the o r i e n t a t i o n i s preserved;

admits no s i n m l a r p o i n t s o t h e r than the

m + 2 logarithmic

.

branch-points described i n (ii)

I n o r d e r t o c o n s t r u c t such a covering s u r f a c e by a method s i m i l a r t o t h a t of Section 41, the following remark i s very important.

,...,

A ) of Km whose v e r t i c e s a r e A 1 jh jl r e g u l a r with r e s p e c t t o t h e mapping c(A) , i f

K(Aj

(pfd

c ( A j )#c(Aj ) P 9

,...,A

A subpolygon

i s s a i d t o be jh

.

Then t h e remark i s given a s the lemma below.

m L 3 1 . Assume : that (43.1)

4%)# c(%+l),

and t h a t t h e mapping

c ( A)

.., m f l

; A&2=Ao)

f

takes a t l e a s t t h r e e d i s t i n c t v a l u e s .

A

s a t i s f d n n t h e conditions: j% these a r c s do n o t i n t e r s e c t each o t h e r i n t h e i n t e r i o r of Km ; these a r c s subdivide Km i n t o subpolygons a l l o f which a r e r e m l a r

t h e r e e x i s t s a family of arcs (a) (b)

(k = 0,1,.

with r e s p e c t t o t h e mappinq

c(A)

.

.

We s h a l l use induction on m If m = l , the three p o i n t s c(Ao), c ( % ) and c(A2) must be d i s t i n c t , and hence the a s s e r t i o n o f Lemma 43.1 i s true. I n general, t h e r e exists a p a i r %-l,&+l such t h a t Proof:

A GENERAL

184

(43.a

c(%-l)

As t o t h e s e t

Case I:

{c(Aj);

BOUNDARY VALm PROBLEM

-

#C(%++

j =0,1,. ..,k-l,ktl,.

..,ni+l}

, there

a r e two cases:

This s e t c o n t a i n s a t l e a s t t h r e e d i s t i n c t p o i n t s ;

This set c o n t a i n s o n l y two d i s t i n c t p o i n t s .

Case 11:

.

I n Sase I , we can complete t h e proof o f Lemma 43.1 by i n d u c t i o n on m

rase 11, t h e f a m i l y of a r c s

$Aj

(jfk-l,k,k+l)

polygor,s which a r e r e g u l a r w i t h r e s p e c t t o

Km i n t o sub-

subdivide

.

c(A)

In

Therefore t h e a s s e r t i o n

of Lemma 43.1 i s t r u e f o r t h e g e n e r a l c a s e .

,!&.

W e c o n s t r u c t e d a covering s u r f a c e

An a p p l i c a t i o n of a l i n e complex.*

R 3 i n S e c t i o n 41. We can d e s c r i b e t h e method i n S e c t i o n 4l by u s i n g a l i n e ccmplex ( o r a t o p o l o g i c a l t r e e ) .

I n S e c t i o n 41, we d i v i d e d t h e

itiemam s p h e r e by t h e s q u a r e with t h e v e r t i c e s

0, 1, l+i and

p r o c e s s y i e l d e d two domains

D1 and D2 on t h e Riemann s p h e r e .

denote t h e s e two domains by

@

Lo, L1, 1

,

L2

1 to

and

L

and

8

l+i , and

l+i t o

i

, respectively.

i

0

to

(See F i g .

,

de sha.11 u s e t h e n o t a t i o n s

W e shall

,

0

u.1.)

order t o i n d i c a t e t h e s e f o u r line-segments on t h e boundaries of

E2

This

Let us denote by

respectively.

t h e f o u r line-segments which j o i n

3

.

i

to In

D1 and

3

3 and

z+o 1

1

respectively,

L2

F i g . u.1.

See, G. E l f v i n g [ S ] and R . Nevanlinna [28; Chapter XI,$Z,pp.

289-2971.

A LINE COMPLEX

187

Using t h e s e n o t a t i o n s , we can denote by t h e symbol

1

* - 2

2

t h e Riemann sphere which i s c u t along

L"L"L 1 2 3 .

The n o t a t i o n a c t u a l l y

.

If we means t h a t t h e two domains D1 and D2 a r e joined along Lo w a n t t o i n d i c a t e t h e f i v e a r c s ( d . 6 1 , w e s h a l l u s e t h e following n o t a t i o n :

(44-1)

L

and L a r e combined i n t o 192 292 S i m i l a r l y , t h e Riemann sphere which i s c u t along

This n o t a t i o n means t h a t t h e two a r c s a single arc

LyL2

-

.

L2,2 " L 1,2

i s denoted by

-

3

2 m

i

1

0

I n S e c t i o n 41, we c o n s t r u c t e d a logarithmic-end along L2,y L1,2 using countably many copies along

L

~

. L If ~we

Si

.

( j = 1 , 2 , . .)

J

attach

S1

"y

h

o f t h e Riemann s p h e r e , c u t

t o s u r f a c e (44.1) along

o b t a i n a s u r f a c e r e p r e s e n t e d by

13 2

3

H

1

-

0

P

%

If we a t t a c h

Sj+l

to

S

j

.'

2

f\

0

ii

i n t h e same way a s i n S e c t i o n 41, we o b t a i n a

s u r f a c e represented by

2 (44.3)

3

2

3

2

p I 0 1 0 I

% Hence, by a t t a c h i n g t h e logarithmic-end t o s u r f a c e (44.1) along we o b t a i n a s u r f a c e r e p r e s e n t e d by

u L L2,2 1 , 2

A GENERAZ, BWNDARY VALUE PROBLEM

188

(44.4)

----

2 I w 30

+*+-

a21 .--

I n this manner, we can r e p r e s e n t t h e s u r f a c e

Ro

o f S e c t i o n 41 by t h e

l i n e complex below:

Note, f o r example, t h a t t h e Riemann sphere c u t along

%

i s r e p r e s e n t e d by

the s p t o l :

- w %,2 L2,2

We const,ructed t h e logarithmic-end along

by a t t a c h i n g

t o suri’%f-:ce (u.1) along “L and then a t t a c h i n g Sj+l t o S L2,2 1 , 2 ? j If we s t o p this t h e way explained i n S e c t i o n 4 1 , where j =1,2,

S1 in

....

process a t

2

3

j =N-1

2

, where

NL2

2

, we 3

o b t a i n a s u r f a c e r e p r e s e n t e d by

2

3

S y u s i n g t h e same i d e a along each of f i v e a r c s

r e p r e s e n t e d by t h e l i n e complex below:

2

( a . 6 ) , we obtain a surface

A LINE COMPLEX

189

N

0

0

N

Let us mite this l i n e complex i n the following form:

(44.6)

0

Denote t h i s surface by

AN

.

hl The surfaces

%

s a t i s f y the following con-

d it i o n s :

(i)

the %

(ii)

B"

(iii)

Ro=

a r e simply connected domains on the surface R,

9

gN+1 i aD

u %

N 4 I n Section

. a,we

remarked t h a t , the homeomorphism from -he pen ;agon

L1u L2u L3 , maps the f i v e line-segments (41.5)e i t h e r onto t h e f i v e a r c s ( 4 l . 6 ) , r e s p e c t i v e l y , o r onto the f i v e a r c s ( 4 1 . 7 ) , respectively. The f i v e a r c s (W.6) are represented by symbol onto the Riemann sphere, c u t along

(44.1).

Similarly, the f i v e a r c s (4l.7)

a r e represented by the symbol

A GENERAL BOUNDARY VALUE PROBLEM

190

-p+ 3

3

2

,I 2-6s sj-mbol rreans t h a t two a r c s W

L

-

291

are combined i n t o a

L

and

191

By a t t a c h i n g s u r f a c e ( 4 4 . 6 ) t o ( 4 4 . 7 ) along t h e s i n g l e
0

;&!+.e

0

0 N

Denote t h i s s u r f a c e by

.

” copies.

The s u r f a c e

%

s a t i s f i e s t h e following

conditi on s : (a;

c%

(5;; Lrj (cj

i <’

i s a closed Riemann s u r f a c e of genus zero; i s a simply connected domain on

3”;

SN admits f i v e a l g e b r a i c branch-points

& l+i ; i ) 5 ; admits no

5es:rii;ed

i n (c)

.

over base p o i n t s

0 , 1, l+i,

s i n d a r p o i n t s o t h e r t h a n t h e f i v e branch-points

mom 40.1

PROOF OF

45.

Proof of Theorem 40.1.

In this s e c t i o n , we s h a l l explain how t o prove of Section

Theorem 40.1 by using t h e covering s u r f a c e 62, Step I:

191

Consider the s u r f a c e

%

0.

of Section &!+.This s u r f a c e i s a

closed Riemann surface o f genus zero, and a l s o a covering s u r f a c e over the

S

Riemann sphere

.

Therefore, t h e r e e x i s t s a r a t i o n a l f'unction

z,(x)

such t h a t p = zN(x)

(45 .I)

i s a conformal mapping from t h e Riemann sphere and

pE;FN

.

Since a conformal mapping from S

parameters, t h e r a t i o n a l function

Let

po

rn

.

(45 - 2 )

%

the e x t e r i o r of N

Step 11:

S contains t h r e e

a2

which i s

not over

have

,

were defined i n Section &!+. L e t us f i x a p o i n t

%

on the surface

ZN(0)

=Po

% .

Z$O)

=1

,

x=cN(p)

AM c%

.

for all

Hereafter, we s h a l l f i x an i t e g e r

NkM,

Y

We s h a l l prove t h a t t h e family

{cN(p); N 2 M ]

%.

L e t us denote such a mapping by

P=&)

This mapping i s uniquely determined by the conditions

(45.7)

Cp(0) =Po

9

(p'(O)>O

%

. for

i s uniformlx

1st < 1

%.

2)

To do this, consider

a conformal mapping from t h e open unit d i s c

onto t h e i n t e r i o r of

M (2

mappings (45.4) are well defined i n

bounded i n each compact s e t i n the i n t e r i o r of

(45.6)

qN i n t h e

We s h a l l f i x t h r e e parameters i n

,

be the i n v e r s e of (45.1).

(45.5)

the

Let

(45.4)

such N

onto

by t h e conditions:

(45 - 3 )

Since

, we

"=2,3,...)

P o q q

where t h e z (x)

a2c%

Since

$ , where xE S

onto

zN(x) contains a l s o t h r e e parameters.

be a fixed p o i n t i n t h e i n t e r i o r of

base p o i n t

S

n

If

i s a compact set i n t h e i n t e r i o r of

nmber (a:

GENERAL BOUNDtlRY VALTE PROBLEM

A

192

r

% , there

exists a positive

such t h a t

O < r < l ,

and

(5)

the s e t

G i s contained i n t h e image of t h e closed d i s c

Is1 I r

(45.Ei by mapping

( 4 5 . 6 ) . The f u n c t i o n s defined by

(45.9)

(NP)

+N(F) = c N ( Q ( s ) )

are holomorphic and u n i v a l e n t i n t h e open d i s c ( 4 5 . 5 ) .

Furthermore, (45.3)

arid ( 4 5 . 7 ) imply t h a t

$N(O) ='

( 4 5 .lo!

(N~M)

=Wf(O)

9

.

r-T

eref fore, by v i r t u e of Koebefs d i s t o r t i o n theorems,

ir, t h e open d i s c ( 4 5 . 5 ) .

a n i f o r m l y bounded i n t h e c l o s e d d i s c (45.8).

i s unifclrmly bounded i n t h e compact s e t Step 111:

{$N(s);N l M ]

This means t h a t t h e family

is

Hence t h e family (CN(p); N)M!

n.

Since m

u %, ,

Ro=

N =2

we ciln s e l e c t a sequence { c N ( p ) ; N223

.

{cN

( p ) ; j =1,2,. .] from The iamily

j SO

(45.12)

that

lim j-tm

6N

j

e x i s t s uniformly i n each compact set on t h e s u r f a c e

R,

.

By v i r t u e of

( 4 5 . 3 ) , the mapping (45.13)

X=C(?)

i s u n i v a l e n t from

Ro

i n t o t h e complex p l a n e .

Note t h a t t h e image

t h e p o i n t a t i n f i n i t y by mapping (45.1) i s i n t h e e x t e r i o r of denote by

A

t h e image of

Ro

by mapping ( 4 5 . 1 3 ) .

connected domain i n t h e complex p l a n e .

*

See, f o r example, R . Nevanlinna [28;

Then A

qN of Let us

i$ . i s a simply

Let

(3.11), ( 3 . 1 3 ) and (3.15) on p . 911.

PROOF OF THEOFEM 40.1

193

P=

(45.4)

be t h e i n v e r s e of (45.13).

Then

i s meromorphic i n the domain

g(x)

A

.

We s h a l l prove t h a t

(45 - 1 5 ) of t h e open set

uniformly i n each comDact subset contain any poles of

.

g(x)

Let

A

fi be t h e image of

, If fi

does nc

. I

by mapping

(45 .4). Since fi i s compact, w e have (45 -16) if

jo

j2jo

for

fi'=BNj

This means t h a t the functions

i s sufficiently large.

(45.17)

fj(x)

=cN

(g(x)) j a r e w e l l defined i n the compact s e t

uniformly i n

5

.

( C f . (45.12) .)

j -+=

uniformly i n

fi

fi , where A

.

S i m i l a r l y , we can prove t h a t

Vo

the closure of co-neighborhood of the

which does not contain any poles of

fi Vo

(45 120)

.

o
.

Let

Then t h e r e efists a p o s i t i v e i n t e g e r j(e)2jo ;

I f j ( x ) -XI < c/8

-

for

(45.21).

g(x)

.

j o i s s u f f i c i e n t l y l a r g e , the functions

(45.21)

The s e t Vo if

i s a compact

c0

Formula (45.18) i s a l s o uniformly v a l i d i n Vo

defined and univalent i n

(b) for

Furthermore,

i s a f i x e d p o s i t i v e number.

z e t (45.19) uniformly i n

(a)

.

j

Denote by

€0

c i e n t l y small. If

fi

l i m f ' ( x ) =1

(45.19)

subset of

(j2jo)

lim f (x)=x j+m j

(45.18)

set

,

x€ij

,

j 2 j ( € );

This means t h a t

6

f.( )

J ?

.

i s suffi-

( j 2j o )

Hence we a r e well

be a p o s i t i v e number such t h a t

j(c)

such t h a t

A SENERAL BOUNDARY VALUE PROBLER

194

- x J < -1s l f ’ ( x ) I 4 j ThereCore, by v i r t u e of Koebels theorem%, f o r each f i x e d Ifj(X)

(45 2 2 : for (45.21).

(x,j) i n ( 4 5 . 2 l ) , t h e r e exists a p o i n t

h.(x) J

i n t h e €-neighborhood o f

x

such tkii%t

(45 -23; b (x)

Since

‘j

f.(h.(x)) =x , J j i s i n t h e €-neighborhood o r

(45.24:

Ih,(x) rl

x

, we

have

-XI < . E

From (45.17) and ( 4 5 . 2 3 ) we d e r i v e

,

h.(x! = 6 ( z N (XI) 3 j

(45.25’

ttnd henze (45.24) i m p l i e s t h a t

(45 .rb:

IC(Z”(X))

-XI

<€

J

C ( zN

‘This means t h a t the f u n c t i o n s conditions (45.26), i f ( x , j ) g(x)

Remark:

S t e p 1.;: t i o n of

.

2mmIa

(45.15)i s

,

x

.

i s a r b i t r a r y , and

E

t h e d e s i r e d r e s u l t (45.15) f o l l o w s

W e s h a l l prove t h a t t h e Schwarzian of

g(x)

x = O i s i n t h e domain

Farthermore

p

0

is n o t a t i n f i n i t y ,

g l ( 0 ) =1

.

Hence, i f

ro

formula

i s a r a t i o n a l func-

f i r s t o f a l l , from ( 4 5 . 3 ) and (45.12) we d e r i v e

Since she base p o i n t of

.

Vo

Since

a g e n e r a l i z a t i o n o f t h e well-known

This means t h a t t h e p o i n t a t g(x)

are w e l l defined and s a t i s f y

i s i n (45.21).

i s uniformly continuous i n

imme5ia t e l y

(x))

j

A

and

6 (p,) = 0 g(0) =po

x = O i s n o t a pole of i s a s u f f i c i e n t l y small

p o s i t i v e numjer, t h e r e s u l t obtained i n S t e p I11 i m p l i e s t h a t

(45.27.’

, x ) = {g,x)

l i m {z,

j

j-tm

u i i f o r m i y i n t h e domain

(45.28)

’ See,

**

1x1 I r 0

f o r example,

*

R. Nevanlinna

See G. E l f v i n g [8; p.311.

.

[ 2 8 ; Koebe Is Theorem on p .

861.

.

PROOF OF mom 40.1

195

Mapping (45.1) i s a conformal mapping from t h e Riemann sphere onto t h e

SN

surface

.

%

Note t h a t t h e surface

over the base p o i n t s

%

t h e boundary of

.

0, 1, lti, i

Since

point i s n o t a branch-point

.

and

has e x a c t l y f i v e branch-points lti

.

These branch-points a r e on

%

qN i s i n the e x t e r i o r o f

of

ZN , this

Thus, by v i r t u e of (45.3), we can conclude

that

z l ( x ) has zeros a t f i v e d i s t i n c t p o i n t s i n x-plane. Denote these N f i v e p o i n t s by 5 ( h = 1 , 2 , 3 , 4 , 5 ) . A straight-forward computation N,h shows t h a t (45.29) where

a

N,h

p,,

and

(45.28) f o r l a r g e

are c e r t a i n constants.

, the El/sN hi j

Since

zl ( x ) # 0 i n domain N

f i v e sequences

...I

j=l,2,

j

(h=1,2,3,4,5)

j'

a r e bounded.

Without loss o f g e n e r a l i t y w e assume t h a t

exist i n x-plane.

Set P.(x) =

5

n

h=l

5

n

( j =1,2,. .)

J

and PJX) =

.

( 1 - A ) 'N: ,h

(LThX)

.

2

h=l Then

l i m P . ( x ) =P,(x) j -tcoJ uniformly i n (45.28). Hence

P.(x){zN ,XI a r e polynomials i n x whose J j degrees a r e n o t g r e a t e r than nine, t h e limit function P,(x){g,x] i s alsc a polynomial i n x of degree not g r e a t e r than n i n e . This proves t h a t

uniformly i n (45.28). Since

{g,x] Step V:

i s a r a t i o n a l function of

We s h a l l prove t h a t

Section 4.0, t h e f'unction

g(x)

A

x

.

i s a c t u a l l y x-plane.

As we mentioned i n

i s represented by a r a t i o of two l i n e a r l y

A GENERAI,

196

BOUNDARY VALUE PROBLEM

independent s o l u t i o n s o f the l i n e a r d i f f e r e n t i a l equation 1

y" t 7{g,xJy = 0

(40.10)

.

{g,x] i s a r a t i o n d function of

Since

, solutions

x

of (40.10) admit

only a f i n i t e number of i s o l a t e d s i n g u l a r p o i n t s i n x-plane which a r e poles of

{g,x]

continuum.

.

If

i s n o t x-plane, t h e boundary of

A

Note t h a t

A

i s simply connected.

Hence

ued a n a l y t i c a l l y across a p a r t of t h e boundary of

since Ro

4 must contain a A

g(x)

.

can be contin-

This i s impossible,

can n o t be extended t o a much l a r g e r covering s u r f a c e over t h e

Hiemann sphere. Step VI:

Ro

We s h a l l prove t h a t

i s a cubic polmomial i n

{g,x]

does n o t admit any a l g e b r a i c branch-points,

zeros i n x-plane and every pole of entire i n

x

,

g(x)

and hence a polynomial i n

conclude t h a t t h e degree of {g,x)

.

i s three.

Since

g l ( x ) does not have any

i s simple. x

.

x

Hence

{g,x]

is

Then from Lemma 40.1 we can Note t h a t

.

m+2=5

This completes t h e proof of Theorem 40.1 f o r t h e s u r f a c e R0

.

For-

mula (45.27) means t h a t the l i n e a r d i f f e r e n t i a l equations 1 y " + ~ { z, x~} y = o (j =1,2, ...) j approximate the d i f f e r e n t i a l equation (40.10) Each of equations (45.30)

( 4 s .30)

.

has e x a c t l y f i v e r e g u l a r s i n g u l a r p o i n t s Riemann sphere. Fuchsian c l a s s . as

h (h=1,2,3.4,5) on t h e j' Hence t h e d i f f e r e n t i a l equations (45.30) a r e of the

N'

Furthermore, the s i n g u l a r p o i n t s

5 N.,h

tend t o i n f i n i t y

J

46. Proof of Theorem 39.1. We s h a l l complete the proof of t h e e x i s t e n c e o f a s o l u t i o n of system (39.11) i n t h e c a s e when (46.1)

m=3

(46.2)

c

0

=o

,

cl=l

,

c2=l+i

,

c3=i

,

.

c =l+i

4

W e have constructed a covering s u r f a c e over t h e Memann sphere i n such Ro a way t h a t conditions (i), (ii) and (iii)of Section & are Isatisfied. W e

have f u r t h e r constructed a function

g(x)

such t h a t

(a)

g(x)

(b)

p = g ( x ) is a conformal mapping from x-plane onto R~ ; t h e Schuarzian {g,x) of g ( x ) i s a cubic polynomial i n

(c)

i s meromorphic i n x-plane; x

.

LOCAL UNIQUENESS

197

W e can assume without l o s s of g e n e r a l i t y t h a t { g , x j =-2[x 3 +alx 2 + a x + a 2 3 Then

g(x)

I.

i s a r a t i o of two l i n e a r l y independent s o l u t i o n s of

y " - [ x 3 + a x2 + a x + a 1 y = 0 1 2 3 Therefore, by virtue of Lemma 40.1,

(46.3)

g(x)+yk as

x-+w i n

%

.

9

(k= 0 , 1 , 2 , 3 , 4 )

,

yk a r e points on the Riemann sphere. F'urthemore, the sets E Y , , Y ~ , Y ~ , Y ~ , Y ~ ~ and E C 0 , ~ 1 , ~ 2 , C433, ~ must be the 981318, s i n c e ( 4 6 - 3 ) implies t h a t yk a r e logarithmic branch-points on Ro Note t h a t % h 2 i s mapped onto 8keh by t h e r o t a t i o n % = w x , where w=exp[.-rri] . Hence 5 where

.

.

Then i t will yo=co tl h a t we have y k = \ follow immediately from condition (ii)of Section &

we can assume without l o s s of g e n e r a l i t y t h a t

.

This completes the proof of Theorem 39.1 f o r t h e s p e c i a l (k=0,1,2,3,4) case given by (46 .l) and (46.2) 47.

.

Local uniaueness.

As we have seen i n Section 42, equations (39.11)

may have many s o l u t i o n s .

This i s not unusual f o r any eigenvalue problems.

In f a c t , Problem 39.1 i s a multiparameter eigenvalue problem i n the complex I n this and the next section* we s h a l l i n v e s t i g a t e t h e d i s c r e t e n e s s

plane.

I n o t h e r words, we s h a l l i n v e s t i g a t e the l o c a l uniqueness of s o l u t i o n s of equations (39.11). To state the problem properly, we must eliminate non-essestial parameters from t h e left-hand members of (39.11). of eigenvalues.

Observe t h a t the boundary conditions (39.10) a r e equivalent t o cp (x+s 9 a,{,

,Il)

(p(x+s,a,52,12) where

+ ck

as

x-+w i n

%

(k=O,

...,mtl) ,

is an a r b i t r a r y but f i x e d complex number. Therefore, by choosing s i n a s u i t a b l e way, we can reduce a1 t o zero. This means t h a t we can s

set (47.1)

a =O 1

i n the left-hand members of (39.11). Observe next t h a t conditions ( i ) and (ii)o f Theorem 39.1 imply the existence of t h r e e d i s t i n c t points

*

.

%, ~ k + ~~ , k + By ~

Sections 47 and 48 are based on I. Bakken [ 2 ] .

r o t a t i n g x-plane

i n a s u i t a b l e way, we can assume t h a t into

2 Ti1,

c 0 ’ c1

and

c2

a r e mutually d i s -

Then a Moebius transformation t a k e s t h e t h r e e p o i n t s

tinct. c

VALUE PROBLEM

A GENERAL BOUNDARY

198

f,

and

0, and

li2

.

1

, respectively.

This allows u s t o e l i m i n a t e

and

T,,

If w e assume t h a t

c = G ,

(47.2)

c o , c1

cl=m,

0

,

c2=1

i: f o i I 2 ; r s f r o m (39 .lo) t h a t 1 ldm o ( x , a ) ’ = C o b ) u(x18J m, 1

c9(x,a,fl,ril) cp ( x 9 a

axi

1 2 ~ ( a ) # O , where

,s 2

7 TI2

Co(a) i s the Stokes m u l t i p l i e r defined i n Section

‘:f. (21.9).)

21,

Se:

l(a2,. ..,am)i

lit=

.

F(x,a2,. .,a ) =

m

Co(0,a2,

. . .,am)# 01 , um,o(x,0,a2,

I

I:0 (0,a2,...,am) m , l ( x , 0 , a 2 ,

--

- 9

am)

...,am ) ’

and

%(a2

,...,am ) = E m

F(x,a2

,...,am )

as

x+m

in

‘k

‘

Recently, I . 3akken [ 2 ] proved t h e following theorem.

THEOFSM h7 .I: For ______ number

(ij

r

%(a2

and -

(a2,

...,am ) 0

& I fo.

there e x i s t s a positive

such t h a t t h e two conditions

,...,am )=%(G2 ,...,-am ) 0

(ii) { a i - a . l < r d J a . =a“ ~j

implv

0

each f i x e d

,

.

( k = 3 , . .,el)

,...,m)

0

Iii.-a.I < r (j=2 J J ( j = 2 , .,m) The constant

..

.

r

may depend on

0

(a2,

.. . , am ). 0

Theorem 4?.l means t h a t t h e mapping % ( a 2 , . . . , a m1 = ck

(47.3)

i s iocally univalent.

Since (47.3) is a n a n a l y t i c mapping, i t s l o c a l

inverse i s also analytic.*

on the .ia-c,a

( co,

.. .

(k = 3 , . ..,m+l)

Therefore, t h e s o l u t i o n o f Problem 39.1 depends

) analytically.

/+8. A sketch of t h e proof of Theorem L7.1.

jC

Note t h a t

See, for example. R. Narasimhan [26; Chapter 5 , Theorem 5, p . 861.

199

A SKETCH OF THE PROOF

F(x,a2,

(48.1)

... , a m ) = F( x,K2, .. .,gm)

f o r every x

i f and only i f

-

.

a. =a ( j = 2 , . .,m) . J j This i s due t o t h e f a c t t h a t t h e Schwarzian of

(48.2)

-2[x

(48.3)

m

+ a2xm-2 + ...+ am]

.

(Cf

F(x,a2,

...,am )

is

. Theorem 40.1 .)

The b a s i c i d e a of t h e proof of Theorem 47.1 is t o derive

(48.4)

x=y

from the equation

(48.5)

F(x,a2

,...,am) = F ( y , a 2 ,... , am . z

U

I . Bakken gave two methods f o r t h i s problem. In t h e f i r s t method, he u t i -

The second method i s based on

l i z e d t h e theory of Nevanlinna-E1fving.s

various p r o p e r t i e s of t h e subdominant s o l u t i o n s Q s e c t i o n , w e s h a l l explain h i s second method.

m,k

(x,a)

.%*

In t h i s

To d e r i v e (48.4) from ( 4 8 . 5 ) , Bakken e s t a b l i s h e d the following r e s u l t .

LEMMA L8 .l: Assume t h a t

(a2,

(ii)of Theorem L7.1.

(i)

e x i s t s a function H(x)

... , a m ) g& m, if r

(a",, . . . , a " m ) s a t i s f y conditions i s s u f f i c i e n t l y small, t h e r e

such t h a t

(I) H i s entire i n x ;

.

(11) F ( x , a 2 , . .,am)=F(H(x),g,,

(m)

IH(x)-x~

-,o

.

x+=

P r o p e r t i e s ( I ) and (111) of

.. .,E m )

H(x)

yield

f o r every x ;

H(x) = x

,

Hence, by using

Lemma 48.1, we can e s t a b l i s h ( 4 8 . 1 ) . This completes the proof of Theorem

47.1. Set

(48.6) and

*

I . Bakken [2; D i s s e r t a t i o n ] .

SQ

I. Bakken [2; Technical Report].

A GEXERAL BOUNDARY

200

VALUE PROBLEM

To prove Lemma 48.1, Bakken derived and u t i l i z e d the following lemma.

Note

that

m: Assume

that

(a2,

...,am)

(a",,

( i ) o f Theorem L7.1.

m, any two of

(48.8-k)

- ,...,Ei ) = + ( a , E ) % ( x , a 2 m

...,a"m )

s a t i s f y condition

t h e following m + 2

%(y,a2

,...,am )

equations (k=0,1,

...,mtl)

are equivalent. Since f o ( a , E ) =Co(a")/Co(a) same.

bakken constructed r

, he

he continued

proved t h a t H(x)

.

The s e c t o r s x

Then,

Since

I$

Uo,U1, .. .,Umtl

I n t h i s way, he could show t h a t t o zero as

i s small i n this compact s e t .

a n a l y t i c a l l y i n the s e c t o r

the asymptotic r e p r e s e n t a t i o n s of Uk

i n a s u f f i c i e n t l y l a r g e compact set i n

I H(x) -XI

by using equation (48.8-k) .

in

(48.5) and (48.8-0) a r e t h e

By assuming condition (ii)of Tfieorem 47.1 with a s u f f i c i e n t l y

x-plane. small

H(x)

, equations

tends t o i n f i n i t y .

I . Sakken [2; Technical Report].

m,k

H(x)

(x,a)

and Q,,,,(x,a)

are v a l i d

cover the e n t i r e x-plane completely. i s e n t i r e and t h a t

I H(x) - X I

For the d e t a i l s of t h e proof, s e e

tends

CHAPTER

2

SUBDOMINANT SOLUTIONS ADMITTING A PRESCRIBED STOKES PHENOMENON*

49.

Main problems.

Id

The subdominant s o l u t i o n s

(49.1)

bm,k(

9

a) = %( a )bm,k+l(

9

a) +

of the d i f f e r e n t i a l

m,k

equation (6.1) admit the connection formulas:

G( )Um,k+2( (k=O,

...,m t l )

,

where

Some p r o p e r t i e s of the Stokes m u l t i p l i e r s Chapter 5.

$(a)

and

$(a)

were given i n

For example,

(49.3) where

(49 *4)

w =exPC-

2n

mt2

i)

and

(49.5 1

v(a)=bl

(a)

.

Zm+l In this chapter, we s h a l l consider t h e following two problems.

m i s odd, and t h a t yo, ...,yWl a r e given m complex numbers al, a m such t h a t

Assume t h a t

complex numbers.

(49.6)

*

Find $(“l9

,am) = yk

( k =O,.

..., .. , m t l ) .

This chapter i s based on Y. Sibuya [4O] and [41].

A PRESCRIBED STOKES PHENOMENON

202

L9.2:

m

Assume t h a t

m

given complex numbers. -2w(a)

(49.7) and

w

(49.8)

\(al,.

i s even, and t h a t complex numbers

-- P

,...,yHl p are al, . . . , a m such t h a t

yo

-

.. , a

(k = 0 ,

) =y

...,m + l ) .

I n t h i s s e c t i o n , w e s h a l l d e r i v e some necessary conditions f o r t h e As i n Section 21, l e t u s s e t

existence of s o l u t i o n s of these two problems.

These matrices s a t i s f y t h e i d e n t i t y

.

(49 .lo)

Sm+l(a)Sm(a). .Sl(a)So(a) =12

We a l s o know t h a t , f o r each fixed

, there

a

.

(Cf.

(21.31) .)

e x i s t s an i n t e g e r

k

such

$hat

i49 .llj

%(a)#O

.

( C f . Theorem 21.4.)

firthemore,

(49 -12)

w - 2 ~ ( a ) + ~for every

a

.

Therefore, we obtain t h e following r e s u l t s . L9.1:

In order t h a t Problem L9.l have a s o l u t i o n , i t i s necessary

t h a t the matrices

s a t i s f y the condition

rm+,rm. ..rlro = i2

(49 -14) L9.2:

.

I n order t h a t Problem L9.2 have a s o l u t i o n , i t is necessarx

that -

(49.15)

P f O

(49 .l6j

ykfO

and t h e matrices

9

f o r some k

,

CASE m = l

ANDCASE m = 2

203

.

s a t i s f y condition (49 .U+)

It should be remarked t h a t we need condition (49.16) o n l y when

50.

Case m = l

=(-l) 2

and case

c0 ( a 1) where

.

2

w=exp{-ni)

3

.

+n

p$ 1

(49 -18)

m=2

.

W e know t h a t , i f

= C ( a ) = C (a ) = l + w 1 1 2 1

(Cf.

Theorem 22.1.)

m=l

,

m=l

, condition

,

Also, i f

( ~ .U) 9 becomes

and i t follows t h a t y

0

=y

1

=y = l + w 2

.

, condition

(49.14) i s a l s o s u f f i c i e n t f o r t h e existence of a s o l u t i o n of Problem 49.1. I n case m = 2 , condition (49.l.4) becomes Therefore, i f

m=l

This i s equivalent t o

Note t h a t , i f (50.2)

and

m=2

, we

Yo=Y2

have 9

.

w=i

Yl=Y3

9

From (50.1), i t follows t h a t

A PRESCFXBED STOKES PHENOMENON

204

(50.3) Conversely, (50.1) follows from (50.2) and ( 5 0 . 3 ) . -2 zero, then p =-1

.

Also, we know t h a t , i f

m=2

1 2 7 1

1 2 v ( a ) = b = -1 a 2 2-iia1

j

is

-in(Lb-L) 2 4-

J-

2n

r(+)

’

’

(Cf. (21.12), (22.10), and Theorem 22.2.) l e t us assume t h a t t h e q u a n t i t i e s

y

,

b ) =C2(al,a2) = 2 e

‘

If one of the

To solve Problem 49.2 f o r

yo,y1,y2,y3

and

m= 2

,

s a t i s f y the condi-

p

tions

(i) p f o , (ii) y o or y , f o

,

and (50.2) and ( 5 0 . 3 ) . r($tb)#m

.

I n case yo# 0 , choose b s o t h a t e-ibn = p and al s o t h a t Co(a , a ) = C ( a a ) = y o ( = y )

Also, choose

1

2 2 1’2 2 Then i t follows t h a t C ( a , a ) = C3(al,a2) = y l ( = y3) W e can t r e a t the 1 1 2 i n the same way. Therefore, i f m = 2 , conditions (49.15), case y,#O

.

.

(49.16) and (49.14) a r e s u f f i c i e n t f o r t h e e x i s t e n c e of a s o l u t i o n of F’roblem 49.2. 51.

Problem L9.1 a t

(51.1)

a=O

.

At

, we

a=O

have

,...,m t l ) .

(k=O

s(0)= l t w

( C f . (23.,!+).)

Therefore, (51.2)

$(a)

#0

(k = 0 , .

.., m + l )

for

(51.3) if

Ro> 0

m M lA 5LL:

1 1 .1

. a * +

bI,

IRo

9

i s s u f f i c i e n t l y small.

a,b,c,a,p,y,t,T,

none o f them i s zero.

&, if

g&

c

be complex numbers.

Assume t h a t

PROBLEM 49.1 AT a = 0

205

i t follows t h a t

(51.5)

a = a ,

Proof:

b=!,

.

c=y

A straight-forward computation shows t h a t

q, and c a r e a r b i t r a r y complex q u a n t i t i e s . Therefore, (51.5) follows from (51.4). This completes the proof of Lemma 51.1. a , b, c , 5 ,

where

LEMMA 51.2: according a s

(49.9).

rk

Let the matrices

m i s odd o r even.

Assume t h a t the

rk

be defined by e i t h e r (49.13) Let the matrices

Sk(a)

or

be defined by

m, if

s a t i s f y condition (49.14).

(49.17) a

i n domain ( 5 1 . 3 ) , the system of equations

(51.6)

%(a)

=rk

(k=O,l,

S,(a)

=rk

(k=O,

...,m+l)

i s equivalent t o (51.7)

...,m-2) .

It i s s u f f i c i e n t t o show t h a t (51.6) follows from (51.7).

Proof:

t h e matrices

%(a)

Since

s a t i s f y r e l a t i o n (49 .lo), we derive

rmtlrmrm-l = s,,(a)sm(a)sm-,(a)

(51.8)

from (49.10), (49.14) and ( 5 1 . 7 ) .

( 51 -9)

By v i r t u e of Lemma 51.1, i t follows t h a t r m = S m ( a ) , rmtl=Smt,(a)

rm-l =sm-l(a)

.

This completes the proof of Lemma 51.2. LEUMA

Assume t h a t

complex v a r i a b l e s . tions

a,(y)

(i)

a,(y)

each

(51.11)

i s odd.

&&

yo

such t h a t

i s holomorRhic f o r Iyj-(l+w)J
a,(y)

m

Then t h e r e exist a Dositive number

( k = 2 , . ..,m)

(51-10) (ii)

m23

(j=O,

satisfies t h e condition apw,l+w,.

..,l+w)

=0 ;

...,m-2)

;

,...,ym-2 be c

m-1

func-

A PRESCRIBED STOKES PHENOMENON

2 6

Proof:

L_

Let us compute t h e determinant

(51.13)

A(a)=

a cO

...

aam at

a=O

.

acm-2

aam

From % ( a ) = C(Gk(a))

,

( c f . ( i i ) of Theorem 21.1),

i t follows t h a t

w

1

w

-4

...

w

w4

...

w

w

-3

-2( m-2) -3(m-2)

m

...

...

-m

...

-m(m-2)

w

w

-2m

w

ac

n K j=2 j

... ... ... 1

"her, A( 0) # 0

-2

1

f o l l o w s from (23.12) of Theorem 23.2.

a=O

Consider t h e system

of e q u a t i o n s

...,

...,

.

C.(0,a2, a =y ( j =O, m-2) J m j Since Z . ( O ) =1+w , and A(0) # 0 , w e can solve (51.15) with r e s p e c t t o J This completes t h e proof o f Lemma 51.3. %2, ,a

(51.15)

...

m '

Assume t h a t

m Q R F ; M 51.1:

t i v e number

E

such t h a t ,

m

m23

if

m+ 2

i s odd.

Then t h e r e e d s t s a posi-

complex q u a n t i t i e s

yo,

...

satisfv t h e c o n d i t i o n s

(51.16)

Juj-(l+w)I

< s

(j=O,

and i f m a t r i c e s (49.13) satisfy c o n d i t i o n

...,m t l ) , (49 .u), a solution

L9.1 is given by

(51.17)

~=~(Y0~.'.'Ym-;?)

(k = 2 ,

...,m)

o f Problem

a =0

PROBLPl 49.2 AT

207

and -

(51.18)

al=O,

where a 2 ( y ) ,...,am(y)

are d v e n i n h

a 51.3.

This theorem follows from Lemmas 51.2 and 51.3. 52.

Problem L9.2 a t

a=O

.

If m i s even, the system of equations

(51.7) i s equivalent t o -2w(a) = P

{

(52.1)

9

.

( k = 0 , . .,m-2)

=yk

.

In order t o solve ( 5 2 . 1 ) , we consider f i r s t t h e system o f equations u(Z) =a )’ “&-l(’) =YoYZh-l

(52.2)

.

,

( h z l , . .,*-l) 1 2

9

where

m-t2 i l o g p

,

(52.3)

m Assume t h a t

m>_4

m

i s even.

m, there e x i s t a p o s i t i v e number ( k = 2 , . . .,m) such t h a t

be complex v a r i a b l e s .

functions

%(y,u)

( i ) each %(y,u)

(52.5)

ak(ltw,l+w

.

Proof:

-

Set

8

, gIJ

_and m - 1

satisfies the condition

,...,1 t w , O )

(iii) a”= ( 0 , a 2 ( y , a ) , , .,am(y,u))

(52 - 4 )

..,Y,-~

i s holomorphic f o r

%(y,a)

(ii)

yo,.

=O ;

s a t i s f i e s esuations (52.2) in domain

CJ

A PRESCRIBED

208

STOKES PHENOMENON

(52.6) (h=l,

afO

...

afo a am

...

afl

aa2

a rl (52.7)

at

A(a)=

a=O

.

...,312 - 1 ) .

aa2

r-om % ( a ) = C ( Ck ( a ) )

,

it fellows t h a t

r k t h e r m o r e , from Lemma 23.1, we g e t

( c f . ( i i ) of Theorem 21.1),

PROBLEM 49.2 AT a = O

209

...

... D=

(52.11)

... ... ... dm - 21, ~

...

... ...

d m-2,Zm+2 l

and

Observe t h a t

,

dl,j=l+w-j

(52 -13)

d2h, j

+

d2h+l, j

+

( h = l , ...,-m-1 1) 2

- j ) w - j(2h-1)

d2h-l, j =

d2h, j = (1+w - j ) w

..

-2jh

(h = 1,. ,=-2) 1 2

This implies t h a t

1

...

-2

...

W

-4

...

W

...

...

1

1

...

1

1

W

V=

W

-z" -m

W

W

...

-2(5+2)

... ...

...

...

-(5+2)

w

w -2m

...

Note t h a t

Hence from (23.34), (52.10) and (52.L$), we g e t (52.15)

A(O)#O

Therefore, equations (52.2) can be solved with respect t o

-m

i n the

. .

210

A PRESCRTBEI3 STOKES PHENOMENON

neighborhood of

LEMMA 52.2:

8=0

.

This completes t h e proof of Lemma 52.1.

AS i n Theorems 7.3

& 21.2, d e f i n e ul, , .. , um

(52.16)

= sm t u l s m-1 Then t h e f u n c t i o n s

i...tum-ls+u

f o ( a ), f , ( a ) ,

a

...t a m-l ( ~ t s ) t a m

m m-1 + (xts) +al(xts)

.

m

. ..,fm-,(a)

defined by (52.6) s a t i s f y t h e

c o n 3t i o n s (52.17)

( j = 0,.

f j (u) = f j( a )

Froof: -

...,um and a .

Let us denote

ul,

their dependence on x

ul(x,a),

by

...,u,(x,a)

to indicate

Set

u(x.a) = (u,(x,a),

(52.183

..,m-2) .

- ..,u,(x,a)) .

rher. ,k

(52.19)

( u ( x , a ) >= U ( W

-k

x,Ck ( a ) )

.

( C f . (21.26).)

Hence

k -k k I \ ( u ( x , a ) ) = C ( G ( u ( x , a ) ) ) = C(U(W x,G ( a ) ) )

(52.20)

= exp{ZEm(w-kx,Gk(a)) }C(Gk(a)) k = e x p { ( - l ) 2Em(x,a)}Ck(a)

.

In t h i s computation, we used formulas ( 2 1 . 7 ) , ( 2 1 . 2 3 ) , and (ii)of Theorem 21.1.

Therefore, we o b t a i n (52.17) f o r

(52.20).

As t o

fo(a)

bl

(52.21)

Zm+l if C

,

from (52.6) and

observe t h a t

m

1

( a ) = - J1 [xm+ 2 %x m-k 2ni C k=l

15 dx

i s a simple closed curve such t h a t a l l r o o t s o f t h e polynomial a r e

contained i n i t s i n t e r i o r . ( 5 2 -22)

bl

v(a) = b

(a)

Hence

(u) =bl

F+l Since

...,m-2

j =1,

(a)

P1

.

( c f . ( 2 1 . 1 2 ) ) , w e o b t a i n (52.17) f o r

j =O

$1 completes t h e proof of Lemma 52.2. Let u s d e r i v e (52.1) from ( 5 2 . 2 ) .

To do this, assume t h a t

.

This

PROBLEM 49.1: THE GENERAL CASE

Yo,...,Ym,2

yj # 0

.

Then, i f

u

and

a r e i n domain (52.4), and t h a t

...,um(x,a)

Let u l ( x , a ) , a"

211

s > 0 i s so small t h a t

denote the q u a n t i t i e s defined by (52.16).

s a t i s f i e s (52.2), i t follows from Lemma 52.2 t h a t

5 i s an a r b i t r a r y complex number.

a l s o s a t i s f i e s ( 5 2 . 2 ) , where

u(5,E) Observe

that

Co(u(5,s)) =exp{2Em(5,E)}Co(g)

(52.23) Since Em(Z,8)

i s a polynomial i n

5

from zero, we can f i n d

5

, and

.

Co(g)

and yo a r e d i f f e r e n t

so that

(52.24.)

c o ( u ( 5 , a )=yo

This means t h a t

a=u(s,g)

.

satisfies (52.1).

Thus we proved the following theorem.

THEOREM 52.1:

Assume t h a t

p o s i t i v e number

E

m24 and m i s even.

such t h a t ,

m, t h e r e e x i s t s

a

if

and matrices (49.17) s a t i s f y condition (49.l4), Problem L9.2 has a s o l u t i m . 53.

Problem L9.1:

I n this s e c t i o n , we s h a l l show t h a t

t h e general case.

condition (49.14) i s s u f f i c i e n t f o r the existence of a s o l u t i o n of Problem 49.1. LEMMA 53.1:

Assume t h a t

s a t i s f y condition (49.14).

m

m23

i s odd, and t h a t matrices (49.13)

Set

,

c 0 =m

cl=o, (53.2) U

k-2

(k=2,

C k = z

Then

co,

...,c

...,m t l ) .

a r e well-defined points on t h e Riemann sphere, and they

s a t i s f y the following two conditions.

(i) c ~ + ~ # (ck =~O ,

...,mtl;

cm t 2-- c 0 ) i

A PRESCRIBED

212

...,cI U t l 3

(ii) t h e s e t (co,

STOKES PHENOMENON

c o n t a i n s a t l e a s t t h r e e d i s t i n c t p o i n t s on t h e

Riemanr, s p h e r e . Proof:

Note t h a t

L_

.

d e t H,#O

This means t h a t

defined p o i n t s on t h e Riemann s p h e r e . follows from p r o p e r t y

(53.3)

'ktl

-2

#O

.

For

.

(k=2,

.

Hence

are well-

m i s odd, p r o p e r t y

(ii)

( i ) , w e s h a l l prove t h a t

To prove

# 'k

Ho = T o

Observe f i r s t t h a t

-

(i)

Since

. . .,c &l

co,

T~

...,m) ,

=yo

and

k>_O , we have

a =1 0

.

This shows t h a t

%+l = r k t l % ' m d heme

' k t l ='k+l7kt

',"herePore,

" k t l ''ktlak

-t % '

-' k t l

-

7

k

0

0

ktl

'%is shows t h a t

'

'k

'k

b

k

c k t l # ~ for

= -det H , # O

,

'k

k=2,

...,m .

Condition (49.14) can be

vrifter: as

Hence

This shows t h a t

T

~

=-- w -~l #

0

.

Therefore,

c f mtl

a

.

This completes t h e

proof of Lemma 53.1.

moRn*!5 3 . 1 :

Assume t h a t

.

s a t i s f y c o n d i t i o n (49 .I&) Proof:

m i s odd, and t h a t m a t r i c e s (49.13) Then Problem L9.1 h a s a s o l u t i o n .

m>_3

By v i r t u e of Lemma 53.1, we can u s e Theorem 39.1 t o f i n d

p l e x numbers

al,

...,am

such t h a t t h e d i f f e r e n t i a l e q u a t i o n

m y " - [ x mt

z "kx

k=l

m-k ly=O

m

com-

PROBLEM 49.1:

THE GENERAL CASE

has two linearly independent solutions yl(x) the boundary conditions

213

and y,(x)

which satisfy

(53.5) Since c0 = a and c1 = O

, we must have

(53.6) where 1 is a non-zero complex number. By utilizing the matrices %(a) defined by (@.lo), we can write the connection formulas (49.1) as

[ld m,k (X,a),bm,k+l(x,a) 1 = [km,k+l(x’a)9bm,k+2(x9a) lsk(a) (k=O,

...,m + l ) .

Set

...,m-1) .

(k=O, Then

and hence

Therefore, (53.8) Set

(53.9) We shall Drove that if k is even, if k is odd, where (53.11)

k=O,

...,m-1 .

214

A

For

k =0

PRESCRIBED STOKES PHENOMENON

, (53.10)

becomes

Also, i t f o l l o w s from (53.1) and (53.7) t h a t

,

t o ( a ) =Co!a) 'To-yo

,

Hence, (53.8) i m p l i e s t h a t

Cl

0

s o ( a ) =1

=1

.

Assume t h a t ( 5 3 .lo) i s true f o r

(53.13)

i

$(a) =

*-'%A

A%A

Let us consider t h e case when k=h+l

, we

.

Co(a) = X Y o

h

This proves ( 5 3 . 1 2 ) .

k = 0,.

. ., h .

Then

if

h

i s even,

if

h

i s odd.

i s even.

If we s p e c i a l i z e (53.10) f o r

get

W e s h a l l prove t h a t (53.14) i s t r u e .

To do this, observe t h a t (53.13)

Since

'h+l =Yh+lThs 'h

'

'htl ='h+laht

'

Therefore, w e d e r i v e

'h

ar++l(a) =yhtl

This proves ( 5 3 . y ) .

from (53.8).

S i m i l a r l y , we g e t

-1 X Chtl(a)

Note t h a t ='yh+l

if

det Hh#O h

.

i s odd.

This completes t h e proof o f (53.10) f o r (53.11). The m a t r i c e s

Sk(a)

f o r (53.11), and s i n c e

satisfy condition

(49.10). S i n c e (53.10) i s true

m - 1 i s even, i d e n t i t y (49.10) becomes

PROBLEM 49.2:

THE GENERAL

215

CASE

.

S&1(a)Sm(a)AHm-1A=12

On the other hand, (49.14) can be written as

=2'

'm+l'mHm-l Hence

Aswl(a)Sm(a)A

*

=relrm .

This implies that

(53.16)

=1

,

c,Ca) =ym

,

.

~ ~ + ~=ydl ( a )

Theorem 53.1 follows from (53.10) and (53.16).

54.

Problem L9.2:

the general case,

method o f solving Problem 49.2.

For

m 2 4 , we have not

found any

Instead, w e s h a l l prove the following

theorem.

THEOFEM 5L.1:

(i)

m24

(ii) yo,

Assume t h a t

m is even; 'ywl , g& p are given complex numbers;

...

(iii) p f ~; (iv) yo#o ; ( v ) matrices (49.17) s a t i s f y condition (49.14)

Then, -

m complex numbers a l , .

there e x l s t

(54 -1)

If

p =0

..,a m

and an i n t e g e r

if

k

is even,

if

k

is odd,

p

(k=O,

f o r some

j

p=O

can be reduced t o the case

.

such

...,mtl) .

then Theorem 54.1 would give a solution o f Problem 49.2.

ever, we s t i l l do not know how t o prove y #O j

.

How-

The case when yo=O but y,#O

To prove Theorem 54.1, l e t us define matrices

Po,

easily.

...,r&l r-

by

A PRESCRIBED STOKES PHENOMENON

216

where

(54.3)

if

k

i s even,

if

k

i s odd,

(k=O,

?k =

...,m) ,

MRtrices ( 5 4 . 2 ) s a t i s f y t h e condition

(54.5)

-

--

rmlfm...rlro = I . ~.

In f a c t , i f we set

’

OP 1

if

k

i s even,

if

k

i s odd,

.

(k = 0 , . ,,m-1)

and

7

I d e n t i t y ( 5 4 . 5 ) follows from (49.L$),(54.7) and ( 5 4 . 8 ) .

Note t h a t

(54.9)

0

=1

.

Hence

THE GENERAL CASE

PROBLEM 49.2:

c Then

0

= a ,

cl=O,

Ok-2 c ='k-2

217

(k = 2 , .

..,m+l)

are well-defined p o i n t s on t h e Riemann sphere, and they

c0,...,cmtl

s a t i s f y t h e condition (54.12)

%+l#Ck

(k=O,

...,W l ;

Cw2=C0)

( C f . the proof of Lemma 53.1.) Furthermore, s i n c e c2 = 1, t h e set {co c ] contains t h r e e d i s t i n c t p o i n t s , 0 , and 1 Hence mtl by v i r t u e of Theorem 39.1, t h e r e e x i s t m complex numbers a l , . .,a and

,...,

. .

two l i n e a r l y Independent s o l u t i o n s

y,(x)

and

y2(x)

m

of the d i f f e r e n t i a l

equation

m m-k yrt- [xm+ c EkX ]y=o k=l

(54.13) such t h a t

Since

c == 0

,

cl=O

where t h e matrices

and

c2=1

, we

must have

Sk a r e defined by (49.9).

o f subdominant s o l u t i o n s by

Let us define another s e t

A PRESCRIBED STOKES PHENOMENON

218

where

f 5.4 .la)

p( a) = m-zv'a)

.

Let

-

(54.19) be

(k=O,

[~k(x),yki~+l(~)]=[~k+l(~) ,?k+2(x)lsk(')

t h e 2onnection formulas f o r t h e subdominant s o l u t i o n s

where

Fmi2 ( x ) =y,(x)

and

7e

3

(x) = y l ( x )

.

?,(x),

-. ...,Y~+~(X),

Then

-

(54.20)

sm (a") =

uhere

(k =0,.

..,m)

and

Then

(54.25) Hence

q-- as

F(x) + ' - 2 -2

x+-

in

ak

(k=2,

...,m + l ) .

219

PROBLEM 49.2: THE GENERAL CASE

Bk = % tk 'k

(54.26 1 Note t h a t %

(54.27)

.

(k=O,

...,m-1) .

- .

.

so(;) =To

Therefore, from (54.27) and (54.26), we can derive

..

,.

s,(a) =rk

(54.28)

(k = 0 ,

i n t h e same way a s i n Section 53.

..,,m-1)

Since t h e matrices

fk

s a t i s f y condi-

sk(Z) s a t i s f y t h e condition .smtl . ( z ) ~ m ( z... ) So(a) =12 ,

t i o n ( 5 4 . 5 ) , and the matrices

we have

A straight-forward computation shows t h a t

-

-

Sm(Z) =rm

.-. smtl(:)

and

..

.

Hence

I 7+1

1 2"+1 B(Z)

=p

(54.29) Q Let

p

9

.

=yk

.

(k = 0 , . ,,m+l)

be an unknown quantity, and define

s

m m-I. x +alx

(54 -30)

+

...+ am-lx

al

,...,am

by

+ am

...+ grn-l ( x + s ) + Ern . a r e polynomials i n s . By v i r t u e o f

= ( x + s p + ZJx+ s ) m - l + The q u a n t i t i e s

al,

...,a m

Lemma 52.2,

we have

(54.31) (k=O,

...,m)

,

and 1

7n

2

(54.32)

Q ~ ( E ) = B(a)

Co(a)Cmtl(a)

.

Then Theorem 54.1 follows from (54.3) , ( 5 4 . 4 ) , (54.29), (54.31), and (54.32). This completes t h e Proof of Tfmxw

Determine

54 .I.

s

so that

Co(a) = y o .

A PRESCRIBED

220

55. of

STOKES PHENOMENON A set

Subdominant s o l u t i o n s a d m i t t i n g a p r e s c r i b e d Stokes phenomenon.

m + 2 n o n t r i v i a l s o l u t i o n s yo,

...,ymtl

of d i f f e r e n t i a l e q u a t i o n (6 .l)

i s c a l l e d a complete s e t o f subdominant s o l u t i o n s of (6.1) ? i f a r e sub:iominant i n

Note ?hat

m+ 2

-

So,

- go,. ..,Sm C l , r e s p e c t i v e l y .

. ..,Swl

solutions

.

y o , . .,ymtl

( C f . D e f i n i t i o n 7.1.)

cover x-plane completely.

For example, t h e s e t o f

i s a complete s e t o f subdominant solu'dm,o, * * ' " b , m t l Fbrthermore, {yo, .,yrtttl} i s a complete set o f subdomi-

..

tions of (6.1).

nant s o l u t i o n s of (6.1) i f and o n l y i f

.

Y,(x) = a , ~ + ~ , ~ ( x , a )( k = 0 , . .,el)

(55.1) khere t h e

4(

are non-zero c o n s t a n t s .

,

If a complete s e t o f subdominant

s o l u t i o n s of (6.1)i s given by ( 5 5 . 1 ) , t h e s e

m+ 2

subdominant s o l u t i o n s

admit t h e connection formulas (55.2)

( k = O , ...,mtl) where i55.3) and

(55.4) Set

and

(55.5)

$=[;

'01

(k=O,

...,ni-tl) .

Then t h e connection formulas ( 5 5 . 2 ) become

Fhthermore, observe t h a t

where (55.8) and t h e m a t r i c e s

(k=O, Sk(a)

...,rtttl)

a r e given by (49.9).

,

S i n c e m a t r i c e s (49.9)

?

221

A PRESCRIBED STOKES PHENOMENON

s a t i s f y i d e n t i t y (49.10), and

( c f . ( 5 5 . 4 ) ) , matrices (55.5)

Tk+2= T0

s a t i s f y t h e condition

... s

SmflSm

(55.9)

0

.

=12

Conversely, l e t us consider m + 2 matrices

;3

4=[;

(k=O,

...,mtl)

such t h a t

(55.11) In this s e c t i o n w e s h a l l prove the following theorem.

mom

5 5 . 1:

(k = 0 , .

qk r

'k

..,m+1)

be d v e n complex numbers s a t i s -

fying t h e conditions t h a t

(i) matrices (55 .lo) s a t i s f y condition (55 .u); f o r some k if. m i s even. (ii) q k # O

,

Then we can f i n d m

complex numbers

esuation (6 .l)has a comDlete set

al,

.

...,am

so t h a t t h e d i f f e r e n t i a l

{yo,. . , Y & ~ }

of subdominant s o l u t i o n s

which admit the connection formulas

(55.12)

[Yk(x) ,yk+l(x) 1 = [yk+l(x) ,yk+2(x)

where

and

Ymt2=Yo

Proof:

-$(a)

= --w

9

*

*

,*l)

0

W e s h a l l f i r s t consider t h e case when m i s odd.

(49.3) .)

(Cf.

Ye3=Y1

(k = '

(k=O,

I n this case,

...,m f l ) .

Therefore,

Also, condition (55.11) implies t h a t mtl (55.13) II r k = - l , k=O Let u s d e f i n e m + 2

(55

complex numbers

ao,

.u)

where

(55.15)

amf2=ao

,

amt3=a1

.

...,a m t l

by the condition

9

A PRESCRIBED STOKES PHENOMENON

222

'70 do t h i s , s e t a I

0

=1

, ( h = 1,

2(h-1)

<

(55.16

...,$( mtl ) ) ,

C120

= (-w)r

QZh

= ( - wa )m yt l

c

1

,

m+l

he^ it is e t s y to v e r i f y (55.14) f o r

- . :t' r1v';'

n

r

(-WY)

i

.

To prove ( 5 5 - 1 4 ) f o r

k=m ,

=-1

kt2

%j&

frm

ak

k#m

55-13) m d ( 5 5 . 1 4 ) . Then,

II (-w-)

r

"k

a

= -(-u)-

'k+2

%=O

.

m

amt2

Also, froni (55.15), i t f o l l o w s t h a t

n

( - w ?k y ) = (-w)mt2=-1

k=O There:'c-e,

.

kt2

(55.17) y i e l d s (55.14) f o r k = m

.

Set

'kt1

Y k = T qk

(k=O,

...,m+l) ,

Then

rk =

(k=O,

...,m t l )

,

= Ti'0 ( c f . ( 5 5 . 1 5 ) ) , m a t r i c e s (55.18) s a t i s f y c o n d i t i o n (49.14) of Lemma 49.1. Condition (49.14) is s u f f i c i e n t f o r t h e e x i s t e n c e of a s o l u t i o n of Problem 49.1. Therefore, t h e r e e x i s t m complex numbers

Since

'YM2

- ,. .., R m such t h a t

a,

223

A PRESCRIBED STOKES PHENOMENON

=Yk

(k=O,

...,m t l ) .

Set

(55.19)

yk (

=

\urn, k (

7

(k = 0,.

a)

..,mtl) .

Then t h e subdominant s o l u t i o n s (55.19) admit t h e connection formulas (55.12).

This proves Theorem 55.1 f o r t h e c a s e m

Let us consider t h e case when m

i s even.

(k = 0 ,

i s odd.

I n this case

...,m+l)

,

where $ ( a ) = l J -2w(a)

.

( C f . (49.3).)

Therefore, mfl

n

k=O m_

2

n

h=O

m 2

n

h=O

Also, c o n d i t i o n (55.11) i m p l i e s t h a t

mtl

n rk=l. k=O L e t u s determine a complex number

(55.20)

p

by t h e c o n d i t i o n

1

It is e v i d e n t t h a t (55 21) Let u s d e f i n e (55.22)

P f O

-

m + 2 numbers

ao,

...,allt+l

by t h e condition

(k=O,

rk = ( - w p %+2

...,mtl)

,

A

221,

PRESCRIBED STOKES PHENOMENON

where

(55.23)

nW2=ao

,

ad3=a 1 ’

To do this, s e t a

0

=a = 1 , 1

( 5 5 .a+) ‘k= (*p ( - 1 ) kr ) k k-2

Then, (55.22) f o l l o w s f o r

(k=2,

.

k#m,m+l

...,m t l ) .

To prove (55.22) f o r

k=m

and

m + l , observe t h a t (55.20) can be w r i t t e n a s “5-1

r

n

2h

,

( - ~ p , a ) =-p

h=O

2h+2

I n t h e same way as we proved (55.14) f o r k = m

k = m ar,d m + l

from (55.23)

, we

can d e r i v e (55.22) f o r

and ( 5 5 . 2 5 ) .

See,

Then

(55.27)

y,#O

€ o r some

k

,

and (k=O,

rk = T ~ + ~ ~ . , T L ~

...,mtl)

,

where

Since

Tk+z = To

(cf

. (55.23)),

matrices (55.26) s a t i s f y (49.14)

.

There-

f o r e , if w e can s o l v e Problem 49.2, we will be able t o complete t h e proof o f Theorem 55.1 f o r even i n t e g e r s

m

.

However, w e have n o t solved Problen

m24 . I n s t e a d , we proved Theorem 54.1. I n o r d e r t o u t i l i z e Theorem 54.1, l e t us assume t h a t y o # O Other cases can be t r e a t e d i n

49.2 f o r

.

A GENERAT., PROBLEM OF G.D. BIRKHOFF

By v i r t u e of Theorem 54.1, t h e r e exist and an i n t e g e r p such t h a t

m

t h e same way. al,.

.. , a m

if

k

i s even,

if

k

i s odd,

225

complex numbers

(k=O,

...,mtl) .

Set if

k

i s even,

if

k

i s odd,

...,m+l) .

(k=O, W

Then if

k

i s even,

if

k i s odd,

and

-k' = {

W

2P

if

k

i s even,

U

w -2P

if

k

i s odd.

k+Z

Hence, i f we s e t

(5529)

?k ( x, = 'kbm,k (

9

(k=O,

a)

...,mtl) ,

the subdominant s o l u t i o n s (55 . Z 9 ) admit the connection formulas

(55.30)

[ ~ ~ : , ( ~ ) , ~ k + ~ ( ~ ) l = [ ~ ~ + ~ ( X( k) =, O ~ ,~'o',mtl) + ~ ( X )9 l ~ k

fm+2=yo

where

(55.31)

and

y&3="yl

.

This means t h a t , i f w e set

.

(k = 0 , . .,m+l)

s k (x, = a g k (

,

the subdominant s o l u t i o n s (55.31) s a t i s f y the connection formulas (55.12)

.

This completes t h e proof of Theorem 55.1.

56.

A general problem of G.D.

Birkhoff.

As i n Section 11, l e t us consider

a system of t h e general form k y' =x B ( 4 Y

(56.1) where

k

9

x i s a complex independent v a r i a b l e ,

i s a non-negative i n t e g e r ,

y i s an n-dimensional v e c t o r , and t h a t t h e components of

B

i s an n-by-n

matrix.

W e assume

B(x) are convergent power series i n x

c o e f f i c i e n t s are complex numbers. W -h B(x) = Z %x (56.2)

h=O

.

Set

-1

whose

A PRESCRIBED STOKES

226

PHENOMENON

Assume t h a t t h e l e a d i n g c o e f f i c i e n t m a t r i x d i s t i n c t eigenvalues

hl,o'

* '

mit,lial;y d i s t i n c t eigenvalues large.

.

.,A

n,o i1(x),

Bo h a s n mutually

Then t h e m a t r i x

B(x)

. . .,An (x) , i f I XI

a l s o has

n

i s sufficiently

Furthermore, t h e s e eigenvalues are convergent power s e r i e s i n

-1 x ,


Set m

c A.

q x j=

(56.3)

h=O F s r each p a i r

56 .I*) :he

-h

(j=1,

,hX

...,n ) .

(j,.C) such t h a t

j f t

,

j,t=1,

...,n

,

e~~ation

rietexxines

d i r e c t i o n s i n x-plane. The t o t d number o f p a i r s s a t i s 1 p ( n - 1 ) . This means t h a t n ( n - l ) ( k + l ) d i r e c t i o n s a r e

2(k+lj

(56.4)i s

r'ying

d e t e i m i n e i through e q u a t i o n s (56.5)

.

Assume t h a t t h e s e d i r e c t i o n s a r e

mutualiy i i s t i n c t , and denote them by (

56, .6)

a r g x = TV

( w = l ?...,

N)

,

where

(56.7)

IJ=n(n-l)(k+l)

.

Assume a l s o t h a t

(56.9) L e t us denote by

(56 .lo)

8 V

TV

5 arg

A s we p a s s from 8v-1 ?.e[ (X

j,o

-

h

t,o

the sector

)xktl]

to

x<

T~~~

SW

, exactly

.

( w =1,. . ? N )

.

one o f t h e q u a n t i t i e s

changes from p o s i t i v e t o n e g a t i v e .

W e denote it by

L e t us go back t o Theorem 11.1 o f S e c t i o n 11. System (56.1) i s a

A GETERAL PROBLEM OF

the matrix

T

227

Therefore, we can apply Theorem 11.1 t o

s p e c i a l case of system (11.7). system ( 5 6 . 1 ) .

G.D. BTRKHOF'F

,

B(x) of (56.1) does not depend on

Since the matrix

of transformation (11.14) i s independent of

G

.

Let us

w r i t e (11.~) as

(56.11)

y=T(x)z

.

of the asymptotic expansion (11.12) are a l s o independent Th I n other words, components of t h e matrices Th a r e constants.

The matrices of

.

8

kt us write (11.12) a s m

T(x) 3

(56.12)

.

Z Thx-h h=O

x tends t o i n f i n i t y i n s e c t o r (11.13). Let u s

This expansion i s v a l i d as

write system (11.10) a s k z ' = x A(x)z

(56.13)

,

where

(56J 4 )

A(x) = 0 hl(x).*

'

O

I

*

Transformation (56.11) takes system (56 .l) t o (56.14). matrix

T(x)

independent o f

arg x=B

8

.

p2

depend on t h e

.

However, the matrices

Th

and

p1

.

Bh The matrix given by

X k Y(x) = T(x)exp{J 5 A(S)d53

is a fundamental s o l u t i o n matrix of system (56.1).

(56.16)

-k-1 A . ( x )= a . ( x ) + r . x +qj(x) J J J

Set (j=l,...,d

where k

(56.17)

a.(x) =

zh h=O j ,hx

-h

(j=l,

...,n )

and (56.18)

rj --' j , k + l

(j=l,

...,n) .

Then matrix (56.15) can be w r i t t e n a s

(56.19)

A(x)

are

These q u a n t i t i e s can be computed a l g e b r a i c a l l y i n

terns of t h e matrices

(56.15)

and

and the two p o s i t i v e numbers

given direction:

Note t h a t t h e

Y(x) =S(x)xR.eXPVXs kN 5 ) d S l 0

9

9

A PRESCRIBED STOKES PHENOMENON

228

and

%e matrix

S(x) admits an asymptotic expansion

( 5 6 .a)

(D

x Uhx-h

S(x)

h =O tends t o i n f i n i t y i n s e c t o r (11.13).

as

x

tih

are independent of

, but

0

the matrix

The matrices S(x)

R

depends on

, 8

A(x)

.

and

Note

t h a t components of the matrix (56.253

P ( x ) =fSkA(S)dS 0

a r e polynomials i n

x

The two q u a n t i t i e s

o f degree P1

k+ 1

and

.

determine s e c t o r (11.13).

P2

As we

remarked i n Section 11 ( c f . the foot-note f o r Theorem ll.l), these t w o q u a n t i t i e s can be given i n terms of eigenvalues of C.D.

Bo ’

I n 1909,

Birkhoff proved the following theorem.

56.1:” matrix (56.26)

For each s e c t o r

a,

, there

exists a f’undamental s o l u t i o n

Yv(x) of the f 01‘91 y V ( x )= S ( x ) x R expcP(x)l V

,

where m

(56.27)

SV(X)

c

UhX

-h

h=O

as

x

tends t o i n f i n i t y i n Sv

.

This theorem can be proved by u t i l i z i n g Theorem 112. Birkhoff elaborated

this theorem f’urther

*

G.D.

.

Sirkhoff [& $6, pp. 466-4691.

A GENERAL PROBm OF G.D.

mom 56.2 :*

BIRKHOF'F

Choose a fundamental s o l u t i o n matrix

Y1(4 =sl(x)x

(56 -28)

R

exp{p(x)]

229

Y,(x)

of t h e form

,

where (56.29)

as

x

tends t o i n f i n i t y i n g1

.

Then we can construct

Y2,.

..,YN

of

Theorem 56.1 i n such a way t h a t the constant matrices

cV =Yw-l(x) -1YJX)

(56.30)

( w = 2,

..., N )

have the following forms:

+

Cw = ln awEv

(56.31)

(w=2,

...,N )

,

where ln i s t h e n-by-n i d e n t i t y matrix; a2, ...,% a r e complex numberg each matrix E i s an n-by-n matrix of zero components except f o r the W

sincrle comonent i n t h e 4, - t h row end j y - t h column which i s esual t o V one. Furthermore, we can construct another fundamental s o l u t i o n matrix

-

%+l(x)

s a t i s f u i n g t h e following conditionst,

(i) YN+l(x)

has t h e form

(56.32) where

as

h=O tends t o i n f i n i t y i n the s e c t o r

x

T1+ 2nIaI'g X < T 2 + 2n ;

(56.34) (ii)

x uhx-h

O3

SN+JX)

(56.33)

t h e constant matrix

( 5 6 -35)

$+I = yN ( )-1%+1 (

has t h e form (56.36)

%+1= l n

%+1

where

+

a~+l%+l 9

i s a complex number, and the matrix

%+1

i s an n-by-n

matrix of zero components except f o r t h e s i n g l e comDanent i n the

row and

j,-

t h column which i s eaual t o one.

independent of t h e choice of

*

Yl(x)

~

G.D.

Birkhoff [4; $6, pp. 463-4669].

.

The matrix

YN+l(x)

"-&

A PFESCRIBED STOKES PHENOMENON

230

Ibis theorem can Se a l s o proved by u s i n g Theorem 11.1. However, w e must a l s o i w e s t i g a t e c a r e f u l l y t h e change o f t h e o r d e r of magnitude o f t h e quantities

.

Re[Xj ,oxkt1]

If we s e t lil(x) =YNtl(e2 n i x)exp(-2niR]

155.3:‘: then t h e m a t r i x

,

s a t i s f i e s t h e c o n d i t i o n o f Theorem 56.2.

Y1(x)

(56.28;, ( 5 6 . 2 9 ) , ( 5 6 . 3 2 ) , (56.33) and (56.34) .) d e n t of t h e choice of

in Theorem 56.2.

( x ) =Yl(e

-Y

-2ni

Tit1

i s i n sector (56.34).

x

Since

YNtl

i s indepen-

can u s e (56.37) i n t h e c o n s t r u c t i o n given

This means t h a t

(56.j8: vhere

, we

Y,(x)

(Cf.

t 56.39)

- -. CNtl

This means t h a t t h e m a t r i x

i n s e c t o r (56.34).

11 = exp{2niR] [ C 2

(56 .LO;

,

Observe t h a t

= Y1(X)C2

YN+&X)

x)exp{ZniR]

... CNt1 I-’

i s t h e c i r c u i t m a t r i x o f t h e fundamental s o l u t i o n m a t r i x singular point a t

X=W

.

Yl(x) f o r t h e The eigenvalues of t h e c i r c u i t m a t r i x o f a

Pmdamental s o l u t i o n m a t r i x

Y(x)

of (56 .l) a t

x = a are independent of

+,he choice of a p a r t i c u l a r fundamental s o l u t i o n m a t r i x

Y(x)

.

S i r k h o f f a l s o proved t h e following theorem.

56.3:* system ( 5 6 . 1 ) n-by-n

Assume t h a t t h e eigenvalues of t h e c i r c u i t m a t r i c e s o f

at

matrix

are mutually d i s t i n c t .

x=m

such t h a t

?(x)

( i ) t h e components of ( i i ) iim 3 x 1

T(x)

are convergent power series i n

i s non-singular;

X+W

( iii )

the transformation

(56 4 1 )

-

w = T(x ) y

reduces system (56.1) &

(56.42)

k’

w l = x B(x)w

,

where

*

S e e , G.D.

Then t h e r e exists an

Birkhoff [ 4 J and

[5].

x-l ;

A GENERAL PROBLEM OF G . D . BIRKHOFF

231

(56.43) The q u a n t i t i e s Let

h

are

..

Y1, . ,YNtl

n-by-n

constant matrices.

be t h e fundamental s o l u t i o n m a t r i c e s o f (56 .l)

which s a t i s f y t h e c o n d i t i o n s given i n Theorem 56.2.

W,W

(56 44) Then t h e s e

N+1

= "Tx) Y"(X)

(v=1,

Set

...,N t l ) .

m a t r i c e s a r e t h e fundamental s o l u t i o n m a t r i c e s of (56.42)

which s a t i s f y t h e c o n d i t i o n s corresponding t o those i n Theorem 56.2. Furthermore, t h e q u a n t i t i e s

R

,

A(x)

and t h e

av

a r e preserved.

We can

assume, without loss of g e n e r a l i t y , t h a t

R

Birkhoff regarded t h e q u a n t i t i e s a

V

, the

c o e f f i c i e n t s of

a s t h e c h a r a c t e r i s t i c q u a n t i t i e s of system (56.1).

q u a n t i t i e s c o n s i s t of (56.46)

n t n(kt1)

n + n(k+l)

+ n(n-1) (kS1)

A(x)

, and

the

The c h a r a c t e r i s t i c

parameters.

Note t h a t

+ n(n-1) (kS1) = n t n 2 ( k + l ) .

Birkhoff a l s o regarded system (56.42) a s t h e canonical form o f ( 5 6 . 1 ) .

If

we r e g a r d t h e c o e f f i c i e n t s of $x) a s parameters, t h e t o t a l number of 2 t h e s e parameters i s n + n (k+l) ( C f . (56.45) .) Based on t h e s e observa-

.

t i o n s , Birkhoff proposed t h e following problem: PROBLEM 56. 1:* Flnd a canonical system (56.42) whose c h a r a c t e r i s t i c q u a n t i t i e s assume p r e s c r i b e d v a l u e s .

His method I n 1913, Birkhoff published a method of s o l v i n g Problem 56.1.** i s based on t h e t h e o r y of s i n g u l a r i n t e g r a l equations of Riemann-Hilbert. The problem of Theorem 55.1 i s similar t o Problem 56.1.

However,

s i n c e t h e form of d i f f e r e n t i a l equation (6.1) i s more r e s t r i c t e d than t h a t o f system (56.42), Theorem 55.1 i s n o t a consequence of B i r k h o f f l s r e s u l t .

The proof of 'Iheorem 55.1 i s e s s e n t i a l l y based on Theorem 39.1 which i s due t o R . Nevanlinna [27].

Theorem 55.1 guarantees t h a t

...,am e x i s t f o r t h e given d a t a qk,rk I all ,...,I am] are s u f f i c i e n t l y s m a l l , t h e

al,

*

(k=O

constants If

smoothness of t h e q u a n t i t i e s

~~

*+

m

,...,m t l ) .

See G.D. Birkhoff [ 4 ; 7, pp. 469-4703. See G.D. Birkhoff [5; 6611 and U . DP. 545-5511.

232

A PRESCRIBED STOKES PHENOMENON

a l , . , . , a m with r e s p e c t t o t h e d a t a can be obtained through t h e i n v e s t i g a t i o n i n Sections 5 1 and 52. I n the general case, such information may be obtained through t h e i n v e s t i g a t i o n i n S e c t i o n s 47 and 48.

CHAPTER 10

SUBDOMINANT S OL UT I ONS OF THE DIFFEFf3"lTAL EQUATION y" - x 2 (x-x,) (X-X,) . .(X-Xm)Y = 0

.

57.

Subdominant s o l u t i o n s .

I n this chapter we s h a l l study asymptotic

behaviors of c e r t a i n s o l u t i o n s of the d i f f e r e n t i a l equation

as

...( x-xm)y=o

yll-X L (X-X,)(X-.,)

(57.1)

1 tends t o i n f i n i t y i n a s t r i p

I ~ . [ x l l 16, , W h I 2 O ,

(57.2) where

is a s u f f i c i e n t l y small p o s i t i v e constant. I n general, a problem of finding asymptotic behaviors of s o l u t i o n s i s divided i n t o two parts: ( i ) construction of formal solutions; (ii) j u s t i f i c a t i o n of the formal s o l u t i o n s . To begin wlth, l e t us consider d2y/dz 2 -P(z,cl c )y=O , (57.3) m 6,

,...,

where

cl,

...,cm

P(Z,Cl,

(57.4.)

rf

a r e complex parameters, and

...,

Cm)

= ( 2-Cl) (2-c2)

. ..(

"Cm)

*

we put

(57.5) the d i f f e r e n t i a l equation (57.1) i s reduced t o (57.3).

Let

%(c)

be t h e symmetric polynomials i n

cl,

...,cm

...,cm ) = x-m+ y ( c ) xm-1 + ...+am-l ( c ) x + am(c) .

P(z,c~, (57.6)

The d i f f e r e n t i a l equation (57.3) has a unique s o l u t i o n

(57.7) such t h a t

defined by

y

=urn(z,a ( c 1)

2.34

SU BDOMIN ANT SOLUTIONS (z,c) ;

( i ) b m ( z , a ( c ) ) i s a n e n t i r e function of

(ii)

urn(z , x ( c ) )

and

U$z,a(c))

admit asymptotic r e p r e s e n t a t i o n s 7

uniformly on each compact s e t i n ity

ir,

...,cm )-space

r

m

The q u a n t i t i e s

as

z

tends t o i n f i n -

.

I a r g Z I < -&n

any closed subsector of t h e open s e c t o r

?heorem 6 . 1 . ) Sence

(cl,

(Cf

.

b h ( a ( c ) ) are symmetric polynomials i n

i s a l s o a symmetric polynomial i n

c

.

b h ( a ( c ) ) i s a homogeneous polynomial i n -k -k -k Lenote ( w cl, w cm) by w c: Then

Hence,

Note t h a t

cl,

.

...,

.

c

...,cm

of degree

h

Y 'ldm,k( z , a ( c ) )

(57.12)

i s a unique s o l u t i o n of (57.3) such t h a t

uniformly on each compact s e t i n

(cl,

...,cm )-space

as

z

tends t o i n f i n -

ity i n any closed subsector o f t h e open s e c t o r

(zf.

Theorems 7.1 and 21.1.)

mials i n

C1'.

..

,c

m '

The q u a n t i t i e s

r m,k

a r e symmetric polyno-

.

SUBDOMINANT SOLUTIONS

235

As we mentioned already, t h e d i f f e r e n t i a l equation (57.1) is reduced

t o (57.3) by t h e change o f v a r i a b l e s ( 5 7 . 5 ) . (57.14)

Y=(xl’.’.,xm)

Set

9

and

(57.18)

Em(px,a(py)) = h E r n ( x , 4 y ) )

.

Thus, we obtain the following r e s u l t . JWE0RF;M 57.1 : The d i f f e r e n t i a l eauation (57.1) has s o l u t i o n s (57.16).

These s o l u t i o n s are e n t i r e functions of

(x,p)

, and

satisfy the asymptotic

conditions

(57 -19)

I

fk(x,h)

-

r =( pw”) fL(X,X)

=(

F

--1

rn9k x m9k[l+ O(x 2)]exp{(-l)k+1hEm(x,a(y))}

.-,

11.

p ~ -Xxprrn’k[(-l)k+l+O(x ~p~~

,

1 _2)]exp{(-l)k+\Ern( x,a(y)) ]

uniformly on each compact s e t i n the domain Pf 0

of p-plane sector (57 -20) where

as

x

tends t o i n f i n i t y i n any closed subsector of t h e open

SUBDOMINANT SOLUTIONS

236

If p > 0

.

in

, the

solution

Hence, we c a l l

f

k

tends t o zero as

fk

x

tends t o i n f i n i t y

a subdominant s o l u t i o n i n t h e s e c t o r

%'

Note t h a t -k -k fk(X,A)=bm(Pwx , a ( p Y ) )

(57.22)

*

In this c h a p t e r , w e shall explain how t o f i n d asymptotic behavior of t h e s e subdominant s o l u t i o n s as variable

58.

x

1 tends t o i n f i n i t y i n s t r i p (57.2).

The

will be r e s t r i c t e d t o a s u i t a b l e domain.

Formal s o l u t i o n s of the a s s o c i a t e d R i c c a t i equation.

L e t us consider

a d i f ' f e r e n t i a l equation 2 y"-x p(x)y=O

,

i s a polynomial i n

x

58.1) where

p(x)

, and

1 is a complex parameter.

In

Sections 58, 59, and 60 we s h a l l d e r i v e asymptotic r e p r e s e n t a t i o n s of solut i o n s of (58.1) with r e s p e c t t o

a

and

x

.

I n such a problem i t i s help-

-I

.

f u l t o r e p r e s e n t s o l u t i o n s i n terms of I P ( x ) ~ Furthermore, a represent a t i o n o f y l / y is derived more e a s i l y than t h a t o f y i t s e l f . Set

(58.2)

u=y'/y

.

Then

(58.3)

U'+U

2

- a 2p ( x ) = o .

This i s a M c c a t i equation. The m .

M c c a t i equation (58.3) has two formal s o l u t i o n s

FORMAL SOLUTIONS

237

(58.6) -n-1

(58.7)

Pn(x, = p ( x )

and t h e a u a n t i t x coefficients.

Proof:

and

g(x)

%(XI

9

is a polynomial i n p , p l , ...,p (n+l) u i t h r a t i o n a l

Let us consider the formal series (58.4). Observe t h a t

238

SUBDOMINANT SOLUTIONS

This means t h a t i f we d e f i n e

Pn(x)

by

-

Qj(x)Qh(x)I

(n>_z)

j th=n

:'his completes the proof of Lemma 58.1. respect t o

where

deg

g(x) denotes

"heref'ore, the q u a n t i t i e s

the degree of Pn(x)

g(x)

with r e s p e c t t o

a r e bounded f o r l a r g e values of -1

So f a r , we have n o t s p e c i f i e d a branch of

(58.4) and ( 5 8 . 5 ) .

function of ( i ) ;Q

Now, l e t us consider

x

in

B

.

(58 .lo)

is a simply connected domain i n x-plane;

Ip(x)(>,B>o

for

8 such t h a t X E O

.

-1 Let us choose a branch of

p(x)'

i n the domain, and assume t h a t

-1 (58.11)

l a r g ~ x I M) ~ ~for~

XE

B

. x

.

~ ( x as ) ~a holonorphic

P r e c i s e l y speaking, w e assume t h a t

( i i ) t h e r e e x i s t s a p o s i t i v e number

x

P ( x ) ~ i n t h e formal s e r i e s i n a simply connected -1

p(x)

s o t h a t we can s p e c i f y a branch of

domain

with

,

deg g ( x ) I (m-1) (n+l)

(58.9)

p(x)

, then

m

is

x

If the degree of

.

,

239

FORMAL SOLUTIONS

M is

where

a p o s i t i v e number.

p o s i t i v e numbers (58.12)

Kn

.

Under assumption (58.10), t h e r e e x i s t

( n = 0,1,. .)

such t h a t

~ p ( x ) - ~ p ~ ( Lx ) K l

~for

x~

5Kn

xEie

and

a

and

(58.13)

IPn(x)l

for

n=1,2,

I n d e r i v i n g (58.13) we used i n e q u a l i t i e s (58.9).

... .

By u t i l i z i n g Theorems

3.3, 4.1 and 5.1 o f Chapter I, we can c o n s t r u c t a f u n c t i o n two complex v a r i a b l e s , (w,x) i n such a way t h a t (i)

g(w,x)

wo

of the

i s holomorphic f o r

(58.14) where

g(w,x)

larg

WI

52M

,

IwI

2wo>

0

,

xE B

,

i s a p o s i t i v e c o n s t a n t , and

(ii) g(w,x)

s a t i s f i e s t h e conditions:

f o r ( 5 8 . u ) and Note t h a t i f

N=1,2, po

...,

where t h e

EN

a r e positive constants.

i s a s u f f i c i e n t l y l a r g e p o s i t i v e number, then

(58.16) for

( 5 8 -17) where

xEB 6,

,

(Im[kl(l6,

,

i s a given p o s i t i v e c o n s t a n t .

R ~ [ X I > _ F J,~ Therefore, i f we s e t

-1

(58 -18)

2 h(x,X) =g(Ap(x) , x )

the f u n c t i o n h(x,A)

(58.17)

.

F'urthermore,

,

i s holomorphic with r e s p e c t t o h(x,X)

(x,k)

s a t i s f i e s t h e conditions:

i n domain

SUBDOMINANT S O L U T I O N S

240

f o r (58.17) and

...,

N=1,2,

where t h e

Notice t h a t

N a r e positive constants. 1 -

-1

-1

hl(x,A) = & ( x ) - l p ~ ( x ) ~ [ a p ( x ) ~ ~ d g ( ~ p ( x ) ~ , x ) /ag(Ap(x)2,x)/ax aw+ and hence the second and t h i r d i n e q u a l i t i e s of (58.15) imply t h e second i n e q u a l i t y of (58.19)

.

Let us s e t

(58.20)

-1 u = X p ( x )2 - p1( x ) - ' p ! ( x )

th(x,A)+u"

.

Then, t h e R i c c a t i equation (58.3) i s transformed t o 1

where (58.22)

-1 H(X,A)

= [hp(x)'- ~ ( x ) - ' p I ( x ) + h ( x , A ) 1 '

+

-

1 [Ap(xI2 - 5 ( x ) - ' p t ( x )

+ h ( x , h ) 12 - A 2P ( x )

-

Since s e r i e s (58.4) is a formal s o l u t i o n o f ( 5 8 . 3 ) , i n e q u a l i t i e s (58.19) imply t h a t

-1 -N

(58.23)

I H(x,A)l I$/ I P ( X ) ~ ~

f o r (58.17) and Set

N=1,2,

(58.uJ

iT=hp(x) v

to obtain (58.25) where

...,

where the

$ a r e c e r t a i n p o s i t i v e constants.

-1 2

-1

-12

v f t 2[Ap(x)'+ h ( x , ) , ) ] v + [ h p ( x )

]V

2

+ L(x,A) = 0

FORMAL SOLUTIONS

-1 -1 ( 58.26 )

.

~ ( x , h=) [xp(x)23 ~ ( x , h )

I n e q u a l i t i e s (58.23) imply (58.27) f o r (58.17).

-1 -N

1 L(x,A) I L CN,ll

.

( N = 2,3,. .)

W(d2(

We s h a l l prove the following lemma.

-2"; v1

F

l t w1t w 2

2w

t

(w1

1 1

t w;)

( 1 t w t w2)

2

1

1

(58.31) i s a s o l u t i o n of t h e R i c c a t i esuation (58.3).

Consequently,

SUBDOMINANT SOLUTIONS

59.

As.mptotic s o l u t i o n s i n a canonical domain as

Iri t h i s s e c t i o n , r e s t r i c t i n g

system (58.28) f o r

t o a s u i t a b l e domain

,

) I m [ h ] )< a 0

tends t o i n f i n i t x .

, where

Re[h]lO

Ig

6,

, we

simply connected, and t h a t

should be

should s a t i s f y an i n e q u a l i t y

p(x)

)p(x)l> _ B > o for

( 58 .lo)

s h a l l study

i s a given posi-

I n S e c t i o n 58, we r e q u i r e d t h a t t h e domain

t i v e number.

xhere

x

A

XELI

,

p i s a p o s i t i v e number.

Furthermore, w e assumed t h a t

M

The domain

{ j8.1; srne r e

i s a p o s i t i v e number.

J3 must be r e s t r i c t e d f u r t h e r

by t h e behavior o f t h e q u a n t i t y

-1

X

Re[ A p ( t ) 2 d t

as

A

tends t o i n f i n i t y .

x = 50 i s c a l l e d a t r a n s i t i o n p o i n t of t h e x = s i s a zero of p ( x ) d i f f e r e n t i a l equation (58.1) of o r d e r k , 0 of m u l t i p l i c i t y k .

m

: A point

j

For example, t h e d i f f e r e n t i a l e q u a t i o n 2 2 y"-x x ( x - l ) y = O

has two t r a n s i t i o n p o i n t s x = O and x = l i s of o r d e r two, wNle t h e t r a n s i t i o n p o i n t

.

The t r a n s i t i o n p o i n t is of o r d e r one.

x=l

x=O A

t r a n s i t i o n p o i n t o f o r d e r one i s c a l l e d a simple t r a n s i t i o n p o i n t .

DEFTNITION 59.2:

A curve

x = S(S)

( 0 5 s < so)

i s c a l l e d a S t o k e s curve

f o r t h e d i f f e r e n t i a l e q u a t i o n (58.1), when (i)

5(s)

or -

t= ;

i s continuous f o r

(ii) s ( 0 ) (iii) FJ(s)

O < s < so

, where

s

0

i s a p o s i t i v e number

i s a t r a n s i t i o n p o i n t of (58.1);

for

{ i v ) we have

s>O

i s n o t a t r a n s i t i o n p o i n t of (58.1);

ASYMPTOTIC S O L U T I O N S

where t h e i n t e g r a t i o n i s taken along t h e curve

213

x = 5 (s)

.

Some examples o f Stokes curves will be given i n Section 61.

D E F I N I T I O N 59 .l: A simply connected domain

0

& I x-plane i s c a l l e d a

canonical domain f o r t h e d i f f e r e n t i a l eauation (58.1), (i) t h e boundam of (ii) t h e i n t e r i o r of

(iii)

n

n c o n s i s t s of Stokes curves; n does not contain t r a n s i t i o n

p o i n t s of (58.1);

i s c o n f o d l g mapped bx

(59.1)

z=I(x)

=sxp ( t )-d t 2

X

0

onto t h e whole z-plane c u t by a finite number of v e r t i c a l s , each of which

i s unbounded, where

xo i s a Doint on t h e boundary o f

n

.

Examples of canonical domains will a l s o be given i n Section 61.

n

B of a canonical domain

c o n s t r u c t a subdomain

so t h a t

B

We s h a l l satisfies

conditions (58.10) and (58.11).

n

Mapping (59.1) take5 t h e domain

conformally onto t h e whole z-plane

c u t by a f i n i t e number of unbounded v e r t i c a l s . c a l c u t s by V

j

L1,

...,4,.

of c e n t r a l angle

(iii) (h# j )

For each c u t

j

i s on t h e b i s e c t o r o f

t h e c e n t r a l angle

6

of

a r e not contained i n

W e assume t h a t

6

V

V V

j ’ j

domain derived by removing

VluV2w.

V

s h a l l construct a s e c t o r

j ’

i s so small t h a t t h e o t h e r c u t s

j -

i s independent of

, we

L j

such t h a t

i s contained i n t h e i n t e r i o r of

(i) L.

J (ii) L

6

Let us denote these v e r t i -

j

.“Vk

.

Let

U

4,

be the simply connected

from z-plane.

(See Fig.

59.1.)

SUBDOMINANT SOLUTIONS

U

Fig. 59.1.

Denote by onto

&

.

IJ

t h e subdomain o f

The domain

0

(58.11) a r e s a t i s f i e d i n I n t h e domain

which i s mapped by (59.1) conformally

i s simply connected and c o n d i t i o n s (58.10) and

B

.

A w e c o n s i d e r t h e two formal s o l u t i o n s ( 5 8 . 4 ) and

(58.5) of t h e R i c c a t i e q u a t i o n ( 5 8 . 3 ) .

We c o n s t r u c t a f u n c t i o n

g(w,x)

which s a t i s f i e s c o n d i t i o n s (58.15) f o r xE B

(59.2) where

w

,

larg[w]l 5 2 M + 2 n

i s a positive constant.

0

, 1.1

Then, i f

Lwo>O po

,

i s a s u f f i c i e n t l y large

-1 p o s i t i v e c o n s t a n t , we can f i x a branch of

(59.3)

for

I

~ ( x )such ~ that

ASYMP"IC

and

SOLUTIONS

245

-12

(58.18')

h(x9-A) =g(-XP(X> , x >

t h e functions

h(x,X)

and

h(x,-X)

9

are holomorphic with r e s p e c t t o

i n domain (58.17). -Finthemore,

h(x,A)

(58.17) and N = 1 , 2 ,

h(x,-X)

I

( 58.19 ' )

f o r (58.17) and

... ,

lh(x,-h)

while

-

s a t i s f i e s conditions (58.19) f o r s a t i s f i e s the conditions

-1 -n

N

X [-XP(X)~]

(x,h)

Pn(x)l < % l h p ( x )

-12 -N-1 1

n=l

9

1

N

1

n=l

N=1,2,.

.. , where the %

a r e p o s i t i v e constants.

We

s h a l l prove t h e following theorem. 59 .k*

If

po

i s a s u f f i c i e n t l y l a r m Dositive number,

the

d i f f e r e n t i a l equation 2 y"-A p ( x ) y = O

(58.1)

admits, i n domain (58.17), two s o l u t i o n s of t h e forms:

-1

{

(59.4) where

F+(x,X)

--1

x

Y =xk,u=P(x)' [1+ F-(x,X)lexp[-AJ

and

-l

x

x

$k,h)lexp[AJ p ( t I 2 d t + J h ( t , A ) d t l

y = y + ( x , A T = ~ ( x'[l )+

,

-l x p(tI2dt+ h(t,-A)dt]

are holomorphic i n domain (58.17)

F-(x,A)

Y

uniformly f o r x E &

Proof:

h

We s h a l l construct

tends t o i n f i n i t y i n the s t r i D

y+(x,A)

constructed i n a similar manner.

.

The o t h e r s o l u t i o n

In order t o construct

-

y (x,A)

F+(x,X)

, we

consider t h e system

*

See M.A.

Evgrafov and M.V.

can be

Fedoryuk [11;$4,Remark 4 . 3 on p . 231.

shall

SUBDOMINANT SOLUTIONS

where

( 58.26 )

1

J.

H(x,h) = [ h p ( ~ ) & ~ (-x ) - l p ' ( x ) t h ( x , h ) ] '

(58.22)

4

-1 2

+ [hp(x)

- a 2P(X)

2

- p1 ( x ) - ' p r ( x )

t h(x,X) 1

9

and hence

-12 -N

I L ( X , X ) 1 IcN-J a p h ) I N = 2 , 3 , . .. , where the

( 5 82 7 ) f o r (58.17) and

are c e r t a i n p o s i t i v e con-

CN-l

stants. We s h a l l reduce system (58.28) t o a system of i n t e g r a l equations by choosing the path of i n t e g r a t i o n i n t h e following manner: point i n in

U

&

.

Then

z = I ( x ) is a point i n

.

U

Let

W e shall join

x

be a

z to

by a curve

(59.7)

c

c(z):

along which

Re[XC(s)]

more p r e c i s e l y , choose

=c(s)

,

c(O)=z

,

O<_s<SCP

i s s t r i c t l y decreasing. PO

I f we do not want t o m a k e

so l a r g e t h a t

In order t o d e f i n e C(z) 1 5-6 i n domain (59.6).

Iarg[A]l

4

unnecessarily l a r g e , we can consider a

p0

domain

(59.6

IIm[il1

The q u a n t i t y

6

16, ,

Re[X12po

,

Iarg[hll

i s t h e c e n t r a l angle of t h e s e c t o r s

1

5,s V

j .

. Consider a

large disc

i n z-plane. Now, i f

parts:

1.

We s h a l l f i x

= 1 I ( x ) l (-R

R and assume t h a t R i s s u f f i c i e n t l y l a r g e .

, we

choose

C(z)

s o t h a t i t c o n s i s t s of two

( i ) an a r c contained i n d i s c (59.8) and

i o r of d i s c (59.8)

.

(See Fig. 59.2.)

If

(ii) a r a y i n the e x t e r -

I z{ = I I(x)( > R , t h e

path

C( z)

i s e i t h e r a r a y i n t h e e x t e r i o r of d i s c (59.8), o r a curve c o n s i s t i n g of

247

ASYMPTOTIC SOLUTIONS

three parts: (i) a line-segment joining z t o the boundary of d i s c (59.8), (ii) an a r c contained i n d i s c (59.8) and (iii) a r a y i n t h e e x t e r i o r of d i s c (59.8). (See Mg. 59.3.) I n any case, t h e length of the arc i n disc (59.8) can be bounded by a constant which depends only on t h e domain U , and i s independent of z If C(z) i s a ray i n the e x t e r i o r of (59.8), then C(z) i s given by

.

~(z): c = z + s e ie

(59.7 ' 1

,

,

(O(s<+=)

where

(59.7")

(arg A+e-rr(

1

1

( 7 -P

i

.

Fig. 59.2.

Fig. 59.3.

Let us define a path i n the domain

$!

by

(59.9)

.

I-1( s) denotes the inverse mapping of I ( x ) The path y(x) j o i n s x t o = i n t h e domain 0 If we define 5 ( s ) by

where

.

(59.10) the path

(59.111

5(s)

y(x)

=I-l(c(s))

y

i s given by

yM:

S=<W

(O<S<+=)

=I-'(I(x)) = x (58.28) t o a system of i n t e g r a l equations

Note t h a t

s ( 0 ) =I-'(c(O))

9

.

. W e s h a l l now reduce system

zL8

SUBDOMINANT SOLUTIONS

E r s t of a l l , we s h a l l c o n s t r u c t a bounded s o l u t i o n o f (59.12) i n domain

(58.17)by choosing

sufficiently large.

p

0

TO do this, note t h a t

l e x p [ - ~ ~ ( ~- (Ix( )s ( s ) ) ) ] 2~1

( 59 J3.l

for

(o<st+=)

.

In f a c t ,

since

Re[AC(s)] i s decreasing along t h e path

.

C(z)

Next, we s h a l l

prove the following lemma.

59,l: In domain (58.17) (59.U+)

where t h e Q u a n t i t i e s

Proof:

K

a r e p o s i t i v e constants independent of

If we change the v a r i a b l e

x

.

c =I(S) , then

5 by

1

(59.16) and

--1

(59.17; Since

p(x)

numbers

c1

s a t i s f i e s condition and

(59.18) where

c2

c,151mclP(S)1 I c , l 5 I m

two p o s i t i v e numbers

where

p(x) and

c4

Ic,l51

for

5E

with r e s p e c t t o

7

x

.

Also,there exist

such t h a t

$1

+c

for

4

IP(S)l >-c51clN ( m t 2 ) f o r

and the f a c t t h a t

(59.21)

t h e r e e x i s t two p o s i t i v e

SE B

.

i s a s u f f i c i e n t l y l a r g e p o s i t i v e number,

5 =I-'(C)

an e s t i m a t e

c3

Icl = I I ( S ) l

Therefore, if R

(592.0)

,

such t h a t

m i s the degree of

(59.19)

(58.10) i n B

and

c5

cEU

i s a positive constant.

Pn(S) =O(S-"-l)

as

5

and .

IcI 2 R

By u t i l i z i n g

,

(58.19)

tends t o i n f i n i t y , we can d e r i v e

249

ASYMPTOTIC SOLUTIONS

where

c6

I n deriving (59.21), w e d s o used

i s a p o s i t i v e constant.

(59.18). Estimates (59.20) and (59.21) imply t h a t

--1 (59.22)

21

lh(S,A)l(p(S)

S=I-l(c)

for

cEU

and

Icl > - R

.

i s a p o s i t i v e number. Therefore by using the 7 f a c t t h a t t h e length of the p a r t of C ( z ) i n d i s c (59.8) i s bounded by a

where

and

c

constant independent o f

f o r a p o s i t i v e number

z

in

, we

a

U

, and

can prove (59.14).

By u t i l i z i n g Lemma 59.1, w e can e a s i l y con-

proved i n a s i m i l a r manner.

s t r u c t a bounded s o l u t i o n w,(x,X) (58.17) i f

po

and

i s sufficiently large.

p o s i t i v e constant

c

w2(x,A)

of (59.12) i n domain

Furthermore, we can a l s o find a

such t h a t

8

Then (59.15) implies t h a t

We s h a l l leave the d e t a i l s t o t h e r e a d e r s . wl(x,h) s o uniformly f o r

Estimate (59.15) can be

x E p as A

,

w,(x,x)

so

tends t o i n f i n i t y i n s t r i p (59.6).

complete the proof of Theorem 59.1 f o r

F+(x,A)= w,(x,x)

,

y+(x,X)

I n order t o

set

+ w2(x,i)

and use Lemma 58.3.

60.

As.wlDtotic s o l u t i o n s i n a canonical domain a s

I n Section 59, w e constructed a subdomain t h a t t h e r e exist two s o l u t i o n s y+(x,h)

&I

and

s a t i s f y t h e conditions given i n Theorem 59.1. i n v e s t i g a t e the asymptotic behavior o f tends t o i n f i n i t y i n behavior of

F+(x,A.)

DEFINITION 60J:*

B

.

and

x

tends t o i n f i n i t x .

n

of a canonical domain y- (x.A)

so

of (58.1) which

I n this s e c t i o n , w e s h a l l

y+(x,l)

and

y-(x.X)

as

x

I n o t h e r words, we s h a l l study the asymptotic

F- ( x , l )

as

A canonical domain

x

tends t o i n f i n i t y i n

n is

B

.

s a i d t o be c o n s i s t e n t ,

or

i n c o n s i s t e n t , according a s t h e v e r t i c a l c u t s i n i t s i m a g e i n the z-plane do, o r do n o t , a l l tend t o i n f i n i t y i n t h e same d i r e c t i o n .

*

This d e f i n i t i o n i s due t o W. Wasow [49].

SUBDOMINAN T SOLUTIONS

250

Set

where

r

i s a p o s i t i v e number.

Then, i f

i s a n c n s i s t e n t canonical domain,

O,S(r)

ari I n c o n s i s t e n t canonical domain,

( S , 8

ea,,h of uhich i s simply connected.

i s s u f f i c i e n t l y l a r g e , and

r

i s simply connected.

If

n

c1 i s

r ) has two connected components

W e s h a l l t r e a t t h e s e two c a s e s separ-

ately.

'I'HEORDl 60.1:"

rf n

i s a c o n s i s t e n t canonical domain,

f3I' (58.i7) N = 1 , 2 ,... , where t h e Praof: bJe s h a l l t r e a t F+(x,A) only.

s t u d i e d i n a s i m i l a r manner.

tim ticjn

$

then

a r e c e r t a i n p o s i t i v e numbers.

The f u n c t i o n

F ( x , l ) can be

I n S e c t i o n 59, we c o n s t r u c t e d a bounded solu-

and w of t h e system o f i n t e g r a l e q u a t i o n s (59.12). 1 2 $ + ( x , A ) i s defined by

The h c -

x,

.

F+:x,X) =w,(x,A) + w 2 ( x , h )

;%eref'ore, we s h a l l s t u d y t h e asymptotic behavior o f tends t o i n f i n i t y i n

&I

.

w1

and

w2

To do this, l e t u s change t h e v a r i a b l e

as

5

in

systen! !59.12) by

c

(60.2)

=:(S)

*

Then,

wl(x,A =

i(

[-2h(S,A)wl(I,h)

+ ~1L ( S , h ) ( w l ( S , h )+ wZ(S,h))

z)

1 -

+ $LCS,X) M S )

2

e x p [ - a ( z - 6 ) 3%

,

(60.33 w2(x,h)

=J-

[ 2h( S ,A )w,( 5 ,A )

c( a )

-$5,

A ) ( ~ ~ (A 5+, w2( I, h )

1 _-

1

- $ 5 , h ) l ~ ( I ) 'dc where

z =i(x)

.

Assilme t h a t a l l t h e v e r t i c a l c u t s

' See

M . A . Evgrafov and M.V.

L1, .. .,$ are d i r e c t e d downwards.

Fedoryuk [ll; Theorems 3 . 3 and 4 . 1 , Remark 4.31

251

ASYMPTOTIC SOLUTIONS

Consider a half-plane

I m [ C ] < r i n C-plane, where r i s a real v a r i a b l e . I n this half-plane, we s h a l l construct a closed s e t O ( r ) which s a t i s f i e s the following conditions: (See Fig. 60.1,)

& ( r ) i s simply connected,

( i ) the i n t e r i o r of

&(r) c o n s i s t s of t h r e e parts: = (-1+i r ) - (is)exp[-ig6] 2 (OlS<+=)

(ii) the boundary of

C

( a ) a ray: (b)

a line-segment joining

(c)

a ray:

where

(-1t i r )

(1+ ir)

to

6 = (1+i r ) - ( i s ) e x p [ i 2 86]

(Ols<S.p)

i s t h e c e n t r a l angle of the s e c t o r s V

6

j *

-1+ir

W e shall denote by S(r)

seen t h a t

O(r)fiU

, and

choose

C(z)

i n 8(r),U

(Cf. Section 59.)

i n <-plane.

It i s e a s i l y

r is a

Furthermore, i f

a r e contained i n

Ll1...,$

For a l a r g e p o s i t i v e

i s simply connected.

r

, we

can

as a ray given by

ie

~ ( z ) :c = z + s e

(60.4)

a(.>

the e x t e r i o r of

i s a bounded set i n <-plane.

S(r),,U

1

1+ir

s u f f i c i e n t l y l a r g e p o s i t i v e number, then

a(r)

9

,

(OlS<+o)

9

where

(60.4 I )

larg

~1

6 p

.

Observe t h a t , if a = I ( x ) E g(r),U

-

, and

~ r

i s a sufficiently large

p o s i t i v e number, then

where

c

i s a c e r t a i n p o s i t i v e number.

By using t h e same method as i n the

proof o f Lemma 10.1 (Chapter 2), we obtain

SUBDOMINANT SOLUTIONS

252

1 w l ( x , a ) 1 I tll ~

-

:L

where

-i s

1

for

( x -2)

a p o s i t i v e number.

Since

x~

I-l(s(r)nu) ,

I-1(8(r)nU)

i s bounded, by

L1 i n a s u i t a b l e manner, we g e t

choosing

I wl(x ,A 1 I El/

( 60.5 ')

x) I

-2

.

f o r (58.17)

In d e r i v i n g (60.5),w e used t h e f i r s t r e l a t i o n of ( 6 0 . 7 ) and estimate (59.22).

Note t h a t t h e e x i s t e n c e of

S e c t i o n 59.

was proved a l r e a d y i n 2 Iwl(x,A)l and Iw2(x,X)l

and w

w1

What we a r e doing i s t o e s t i m a t e

Now, Sy u s i n g t h e second r e l a t i o n of (60.3) and (60.5) we can a l s o d e r i v e

(60.5 'I

f o r (58.17)

Iw2(x,h)l ( Z , I A I ( X ) ~ - ~

.

.

I n d e r i v i n g ( 6 0 . 5 ' ) , we L1 as w e want, s i n c e t h e second r e l a t i o n o f (60.3)

'de must choose a s u i t a b l e p o s i t i v e number

-an mcdify t h e p a t h

-

C( z )

does not c o n t a i n e x p o n e n t i a l terms.

We l e a v e t h e d e t a i l s t o t h e r e a d e r s .

By i n d u c t i o c , we can prove t h a t

f o r (58.17) and

N=1,2

,...,

where t h e

5

a r e certain positive constants.

This completes t h e proof of Theorem 60.1 f o r

F+(x,h)

.

S i m i l a r l y , we can prove t h e following theorem.

m 0 H F M . 60.;2:* B

A

for

Assume t h a t

n

i s an i n c o n s i s t e n t canonical domain.

be two a r b i t r a r y r e a l numbers.

(58.17)

N=1,2,.

.., where

p o s i t i v e c o n s t a n t s which depend on

m,

and

:t must be remarked t h a t t h e c o n s t a n t s

constants

f,

depend on

B

.

hold i n t h e same domain ( 5 8 . 1 7 ) . e a s i e r than t h a t o f Theorem 60.1.

. Evgrafov

TN -

depend on

A

, while

the

Note a l s o t h a t estimates (60.1)

This i s a d i f f e r e n c e between a c o n s i s t e n t

and an i n c o n s i s t e n t c a n o n i c a l domain.

See M.A

are certain

, respectively.

B

This means t h a t t h e two estimates (60.6) do

n o t hold i n a common unbounded domain.

'~

&

the quantities

A

Let

The proof of Theorem 60.2 i s much

I t is l e f t t o t h e r e a d e r s as an exercise.

and M.V. Fedoryuk

25 3

EXAMPLES

61. Ejcamules of Stokes curves and canonical domains.

I n order t o f i n d t h e

general behavior of Stokes curves, the following observation i s very help-

ful. Assume t h a t

i s holomorphic i n a simply connected domain 8

g(x)

,

and s e t X

(61.1)

I(x)

=J g(S)dS , X

where

x

0

0

i s e i t h e r i n Q o r on t h e boundary of

i n t e g r a t i o n i s taken i n

(61.2)

B

.

Q

, and

the path o f

Assume a l s o t h a t a smooth curve

x=x(t)

s a t i s f i e s t h e condition R e [ I ( x ( t ) )3 =constant,

(61.3)

i t follows t h a t d

(61.?)

z(t

where

denotes t h e .conjugate complex.

This means t h a t (61.2) s a t i s -

f i e s the d i f f e r e n t i a l equation

-

&/cia = i g ( x )

(61.8)

The d i f f e r e n t i a l equation (61.8) represents a system of two autonomous

d i f f e r e n t i a l equations, i f we take t h e real and imaginary p a r t s of account.

x into

I n other words, curve (61.2) i s an o r b i t o f the system of d i f f e r -

e n t i a l equations (61.8).

On t h e o t h e r hand, i f

x = x ( s ) i s an o r b i t of

(61.8) , then --I(x(s)) d ds

=p(x(s))dx(s)/ds=ilg(x(s))12

and hence R e [ I ( x ( s ) ) ]= c o n s t a n t .

,

SUBDOMINANT SOLUTIONS

251,

Transition points of t h e d i f f e r e n t i a l equation

a r e c r i t i c a l p o i n t s of t h e autonomous system

-1

dx/ds =

(61J G )

im2,

and Stokes curves of (61.9) are t h o s e o r b i t s of (61.10) which s t a r t from o r Therefore, t h e t h e o r y of dynamical systems

end a t t r a n s i t i o n p o i n t s .

defined on two-dimensional manifolds can be u t i l i z e d i n t h e s t u d y o f Stokes curves .% I n o r d e r t o i n v e s t i g a t e t h e l o c a l behavior of Stokes curves of (61.9), x = 50 . can be w r i t t e n

we s h a l l f i r s t consider a neighborhood of a t r a n s i t i o n p o i n t Assume t h a t

x=<

0

i s of o r d e r

m

.

I n this c a s e ,

p(x)

as p ( x ) = (x - S,)ms(x)

(61.ll) where

(61. l z j

q(x)

i s holomorphic i n a neighborhood o f q(So)

#o

x = 50

, and

-

W e are i n t e r e s t e d i n t h e o r b i t s of (61.10) which end a t

x=so

.

Such

o r b i t s are given by

W e can write t h e i n t e g r a l of

where

p ( ~ as ) ~

~ ( x )i s holomorphic i n a neighborhood of -1

x=<

a '

and

2 2 cp(so) = ~ q ( s o )

Hence

*

The o r b i t s of (61.10) a r e c a l l e d t h e t r a j e c t o r i e s of t h e q u a d r a t i c diff-

.

e r e n t i a l -p(x)dx2 ( C f . J . A . Jenkins [20; Chapter 3, pp. 27-43].) M.A. Evgrafov and M.V. Fedoryuk [ll] u t i l i z e d t h e t h e o r y of t r a j e c t o r i e s of q u a d r a t i c d i f f e r e n t i a l s i n t h e s t u d y of Stokes curves and canonical domains. See, a l s o , F.W.J. Olver [30], W. Wasow [ 4 9 ] , and J . Heading [15].

EXAMPLES

as

r

tends t o zero, where

25 5

, and

r=Ix-5,1

(61.15) These m + 2 Stokes curves divide the neighborhood o f t h e t r a n s i t i o n point x=s

0

into

m+ 2 s e c t o r s each of which i s a hyperbolic sector.*

(See

Fig. 61.1.)

/

c2

m=l

Fig. 61.1. I n ord r t o i n v s t i g a t e the behavior of Stokes curves a s

x

tends t o

i n f i n i t y , l e t us assume t h a t

(61.16)

p(x) = xrn(l+ O(x-l)

and

(61.17)

*

-1 p ( x )2 - x

1

,

1

P (1+0(&

For t h e d e f i n i t i o n of hyperbolic, parabolic, and e l l i p t i c s e c t o r s , see P. Hartman Chapter V I I , pp. 161-1741.

[u;

256 as

SUBDOMINANT S O L U T I O N S

x

m i s a positive integer.

tends t o i n f i n i t y , where

The o r b i t s of

(61.10) a r e given by

1 -

x

Re[J ~ ( S ) ~ d=sc ]o n s t a n t . X

0

I t i s e a s i l y v e r i f i e d t h a t t h e neighborhood of

m+ 2

e l l i p t i c s e c t o r s by t h e

i s divided i n t o m t 2

directions

2k+l, a r g x =me2

(61.18)

=-

x

(k = 0 ,

.. .,mtl) .*

This means t h a t Stokes curves tend t o i n f i n i t y i n one of d i r e c t i o n s (61.18). Note t h a t d i r e c t i o n s (61.18) a r e the boundaries of

sectors

So,.

..,

( C f . (7.6) .)

which were defined i n Section 7.

8mtl

m+ 2

in the study of the g l o b a l behavior of Stokes curves, we can u t i l i z e

Ekamples a r e the

various methods i n the theory of d y n d c a l systems.

Poincare-Bendixson Theorem and the theory of i n d i c e s of closed curves .** I n p a r t i c u l a r , the lemma below i s very u s e f u l .

LEMMA 61.1-

0 be a bounded simply connected domain i n x-plane,

l e t i t s boundary

- 61 & B

y

be homeomorrhically equivalent t o a c i r c l e .

the closure of

with no poles on

6,

y

.

.

Assume t h a t a function

p(x)

Denote

i s meromorphic on

Set 1

If

Re[B(x) 3

i s constant on y

, then

p(x)

has a t l e a s t two poles i n

6,,

every pole being counted according t o i t s m u l t i p l i c i t y .

-1 Proof:

of t h e boundary single-valued. p a r t of

y

y

i n the domain

...,5,

L e t ,S,

This branch i s n o t n e c e s s a r i l y

i n the neighborhood of each zero

61.2.)

I f we go around the curve

In g e n e r a l , i f

p(x)

t h e neighborhood of tors.

.

on

p(x)

'j

so t h a t we have a smooth closed curve

B

ic*

&I

be the zeros of

domain

*

~ ( x )i n~ t h e neighborhood

To begin w i t h , l e t u s f i x a branch of

y

.

by a smooth a r c i n the in

.

(See fig.

i n the counter-clock-wise sense, i t s

has a pole o f order g r e a t e r than two a t

5,

Replace a

x=5

0'

is divided i n t o a f i n i t e number of e l l i p t i c sec-

(See J . A . Jenkins [20; Theorem 3 . 3 , p.301.)

See, f o r example, E.A.

Coddington and N. Levinson [7; Chapter 16.61.

and M.V. Fedoryuk [ll; $2, Lemma 2.3, p . 7 1 , and J.A. Jenkins [ZO; L e m m a 3.2, p.361. xx*x C f . M . A . Evgrafov

EXAMPLES tangent makes a complete o f the v e c t o r f i e l d

2n

--

ip(xf2

Now, l e t us compute t h e r o t a t i o n

rotation.

, where -

By v i r t u e o f t h e assumption t h a t

257

denotes the conjugate complex.

Re[B(x)] i s constant on y

, the

differ-

ence between r o t a t i o n s of the tangent and the vector f i e l d i s observed only i n the neighborhood of the zeros of nonnegative i n t e g r a l multiple of

.

p(x)

n

.

Such a difference i s always a

This i s due t o the f a c t t h a t the

neighborhood o f a t r a n s i t i o n p o i n t i s divided i n t o a f i n i t e number o f hyHence, i f w e go around the curve

perbolic s e c t o r s .

7

i n the counter-

clock-wise sense, the vector f i e l d makes a r o t a t i o n of an angle

(2tn)n

-1 where

n

i s a nonnegative i n t e g e r .

t i o n of an angle

-(2+n)n

.

and

, where

N

N-P=-(2tn)

This means t h a t

,

~ ( x makes ) ~ a rota-

Hence

i s the number of the zeros of

i s the number of the poles o f

p(x)

in

completes the proof of Lemma 61.l.

A

.

Thus we g e t

p(x) PL 2

.

, and

P

This

-

c

Mg. 61.2.

The following lemma i s a c o r o l l a r y o f Lemma 61.1.

LEMMA 61.2:*

sf

holomorphic i n

of

61

&I

61

i s a simply connected domain i n x-plane,

, then

o r a transition point i n

.

I n this statement, t h e key assumption i s t h a t

*

p(x)

Q

any o r b i t of (61.10) tends t o e i t h e r t h e boundary

p(x)

does not have any

See M . A . Evgrafov and M.V. Fedoryuk [ll; $2, Lemma 2.5, p.81.

25 8

SUBDOMINANT SOLUTIONS

poles i n

g

.

A s i l l u s t r a t i o n s , F i g u r e s 61.3, 61.4, and 61.5 e x h i b i t t h e general

behavior of Stokes curves f o r t h e c a s e s when ( i ) P(X) = ( x - xl) ( x - x2)

,

(ii) P(X) = ( x - xl) ( x - x2) ( x - x 3 ) ( x - x4) ( x - x 5 ) and

,

EXAMPLES

259

Fig. 61.5. To construct canonical domains, t h e following lemmas a r e u s e f u l .

LENMA 61.2:” y be a closed r e c t i f i a b l e Jordan curve, and denote by & ( y ) t h e domain bounded by y Let p(y) be the closure of 9(y)

.

. -

c

Assume t h a t

f(z)

i s continuous i n

number of p o i n t s of

on

&(y)

and maps

y

, and

9(y)

, holomomhic

.

on y

one-to-one

f(z)

w:

p(x)

be a polynomial i n

x

Assume t h a t

Stokes curves i n i t s i n t e r i o r ,

than

, and

If

&

1

i e a p o i n t on the boundary of

See, f o r example, A . I . Markushevlch

let

&

be a simply

does not contain any

g is conformally mapped e i t h e r onto a

half-plane o r a strip i n 2-plane by the mapping

*

.

i s bounded by t h e Stokes

&I

curves o f t h e d i f f e r e n t i a l equation (61.9).

y

Thus we obtain the

.

connected domain i n x-plane.

where xo

i s univalent

~ ( y ) onto the domain bounded by the Fmaae of

We can use Lemma 61.3 even on the Riemann sphere. following result

except a t a f i n i t e

9

.

[a; p.1201.

SUBDOMIN ANT SOLUTIONS

260

The shaded domains i n F i g . 61.5 a r e examples o f domains which a r e mapped ontc s t r i p s . Jis i l l u s t r a t i o n s , examples o f c a n o n i c a l domains a r e given by Figures

61.6, 61.7,61.8, and 61.9.

These f i g u r e s a r e based on F i g . 61.5.

canonical domains given by Figures 61.6 and 61.7 a r e c o n s i s t e n t , others are inconsistent.

A

k

/

-a-

- bFig. 61.6.

I I

I I

1 \

f

\

\ \

/

\

/

\

/

\

/

I

\

\

Fig. 61.7.

The The

ASYMPTOTIC REPRESENTATIONS

26 1

/

1

/

I

1

I

f

/

F i g . 61.8.

I

E

N g . 61.9.

62.

Aa.mptotic r e p r e s e n t a t i o n s of subdominant s o l u t i o n s a s h

tends t o

Let us go back t o t h e d i f f e r e n t i a l e q u a t i o n (57.1).

W e shall

infinity.

use t h e same d e f i n i t i o n s and n o t a t i o n s as i n S e c t i o n 57. is restricted t o a s t r i p (62 -1)

IIm[Xll 26, ,

The p o s i t i v e number (62.2)

2p0

.

i s supposed t o be s o s m a l l t h a t

1

6

is a preassigned p o s i t i v e number.

fk(x,h) of (57.1) $tisfy

Then, t h e solu-

t h e asympto%ic c o n d i t i o n s (57.19) u n i -

formly on each compact s e t i n domain (62.1) as sectors

A

l u g 11 5x6

f o r ( 6 2 . 1 ) , where tions

6dpo

RehI

The parameter

x

tends t o i n f i n i t y i n t h e

SUBDOMTNANT SOLUTIONS

26 2

s a t i s f i e s t h e asymptotic c o n d i t i o n

--1 uniformly on each compact s e t i n domain (62.1) as

x

tends t o i n f i n i t y i n

the sec tor

(62.5 Ey u s i c g t h e s p e c i a l c a s e when

(62.6)

p ( x ) = (x-x,)

(x-x3) (x-x4 ) (x-x5 ) (x-x,)

(X-X,)

,

w e s h a l l e x p l a i n how t o d e r i v e an asymptotic r e p r e s e n t a t i o n o f h

W e assume t h a t t h e

tends t o i n f i n i t y ,

(62.7)

x <x <x <x <x <x

6

4

5

L e t us denote by

i2+

3

2

and

n-

x

The domain

t h e t w o c a n o n i c a l domains given by

no=n,./ n- , so=n+,,n-

so

no

i s shown by Fig. 62.1.

.

The two canonical domains by t h e mapping

&

and

Set

. The shaded domain i n F i g . 6 2 . 1 i s

n-

are consistent.

1 p(5I2d5 X 1

Their images

x

i62.9)

z = I(x)

=s

have a s i n g l e v e r t i c a l c u t .

-1

Let us choose t h e branch of

-1

(62 .lo)

p(x)2>o

for

Then, mapping (62.9) t a k e s

(62.11)

Re[z]>O

So

as

1 ’

F i g . 61.6-a and Fig. 6 1 -6-b r e s p e c t i v e l y .

(62.8)

j

fo(x,h)

are real numbers, and t h a t

x> x

1 ‘

onto t h e h a l f - p l a n e

.

H

M g . 62.1.

P ( x ) ~s o t h a t

26 3

ASYMPTOTIC REPRESENTATIONS

Let us construct a subdomain

of t h e canonical domain

61+

n+

same way a s we constructed a subdomain 61 o f the canonical domain Section 59.

I n Section 59, w e covered the v e r t i c a l cuts

the sectors

V1,

...,Vk

t i v e number

6

a s the c e n t r a l angle.

of c e n t r a l angle

t h e r e i s only,one v e r t i c a l c u t . o f the canonical domain

n-

.

6

.

L1,.

..,$

i n the

n

in

by

W e use the preassigned posi-

Note t h a t , i n the present case,

Similarly, we construct a subdomain The domain

0

-

- i s shown by Ffg. 62.2.

I n the domain

-

(62.12)

IIm[ll1

w e construct a holomorphic function

5b0 , ho( x , X )

Re[X12po

,

s a t i s f y i n g t h e asymptotic

conditions

f o r (62.12) and

N=1,2,

..., where the

thermore,

is a formal s o l u t i o n o f the R i c c a t i equation 2 u ' t u 2 - X p(x) = o

.

( C f . Lemma 58.1.)

Fig. 62.2.

a r e p o s i t i v e constants.

Fur-

SUBDOMIN AN T SOLUTIONS

264

-1

1 -

Note t h a t w e f i x e d t h e branch of The f u n c t i o n

p(x)

2

so t h a t

p(x)

2

>0

for

X>

x1-

.

can be c o n s t r u c t e d i n t h e same way as we c o n s t r u c t e d

ho(x.h)

i n S e c t i o n 58.

h(x,X)

Ey v i r t u e o f Theorems 59.1 and 6 0 . 1 , t h e r e e x i s t s a s o l u t i o n of (57.1)

o f t h e form:

+s

X

y = p ( x ) '[lt F(x,X)]exp{-XI(x)

(62.14)

ho(t,h)dt]

,

co

xhere

--..

(i: p(xl (iii

--1

1

'

i s t h e branch such t h a t

p(x)

'>

for

0

x > x1

,

F(x,X) i s holomorphic i n t h e domain x€A+

( 6 2 .lj)

,

,

IIm[XlI160

,

Re[h12~,

and s a r ; i s f i e s the c o n d i t i o n s

ioz.:o! f c r !62.15) and

... , t h e

N=1,2,

quantities

$

being p o s i t i v e c o n s t a n t s ,

F i n ii

(iii) ',he path of i n t e g r a t i o n i n t h e right-hand member of (62.14) i s a

ciirve j o i n i n g

$03

to

in

x

at

.

(See F i g . 62.2.)

S i m i l a r l y , w e can

c o n s t r x t a s o l u t i o n of (57.1) o f t h e form:

-1 (62.141 ,

+s

X

y = p ( x ) 4[1+?(x,X)]exp[-hI(x)

ho(t,h)dt)

,

m

where

--1

( i f )p ( x )

i s t h e branch such t h a t

--1 p ( x ) '>

0

for

x>xl

,

(iil) T ( x , h ) i s holomorphic i n t h e domain (62.15 I:

xE9-

)Im[h]l16,

9

Re[X12Po

ani s a t i s f i e s t h e c o n d i t i o n s

(62.16' ;. for (62.15')and (iiiI )

N=1,2

,..., and

t h e p a t h o f i n t e g r a t i o n i n t h e right-hand member o f ( 6 2 . 1 4 ' ) is a

curve j o i n i n g

in

B-

'Theorem 59.1, we must choose

po

+rr,

to

x

.

(See M g . 62.2.)

In order t o use

sufficiently large. The two s o l u t i o n s ( 6 2 . 1 4 ) and (62.141) of t h e d i f f e r e n t i a l equation

( 5 7 . 1 ) a r e subdominant i n t h e s e c t o r

265

ASYMPTOTIC REPRJEXNTATIONS

So: l a r g

s i n c e t h e image of (62.11).

so

<=, n

XI

by mapping (62.9) i s t h e p o s i t i v e half-plane

.(A)

Therefore, t h e r e e x i s t s a function

X

of

such t h a t

.

c ( X ) [ l + F ( x , h ) ] = l + “Fx,h)

c ( h )L 1

By v i r t u e of (62.16) and (62.161), w e g e t

, and

hence

F(x,h) = F(x,A)

(62.17) for

xE&+/\fJ_ Define a function

I I m h I l ‘-b0

9

Fo(x,h)

by

i

F(x,X)

Fo(x,h) =

(62.18)

,

for

x E &+

for

~ € 8 -

.

~

F(x,X)

Then

Re[Xl)_Po

9

i s holomorphic i n domain (62.12), and

Fo(x,h)

I F0(xJ )I

(62.19) f o r (62.12) and

N=1,2

hI(x)

,... .

I

Furthermore, 1

--

y = y o ( x , A ) = p ( x ) 4 [1+ Fo(x,h)lexpC-hI

(62.20)

i s a s o l u t i o n of (57.1),where

--1 p(x) ‘>

0

for

i n t e g r a t i o n i s a curve joining

+-

x

in

to

The two s o l u t i o n s y o ( x , h ) and tor

So

.

fo(x,X)

Hence t h e r e e x t s t s a function

(62.21)

f o ( x , O =c(x)Yo(x,a)

c(h)

, and

x > x1 &+“

.

B-

the path of

(See Fig. 62.2.)

a r e subdominant i n the secof

such t h a t

Z

.

, then

(62.19), (62.20) and (62.21) Will provide an asymptotic representation of fo(x,X) f o r x € & + ~ & a s h tends t o

If we can find

c(h)

i n f i n i t y i n s t r i p (62.1). in

So

, and

-

We can f i n d

c(X)

by l e t t i n g

u t i l i z i n g the asymptotic p r o p e r t i e s of

+.

tend t o

yo(x,X) and

+W

fo(x,l)

The d e t a i l s a r e l e f t t o t h e readers.* In order t o summarize t h e method, again by using case (62.6), we conThis s o l u t i o n i s subdominant i n t h e domain s i d e r t h e s o l u t i o n f2(x,X)

as

x

tends t o

x

.

S2

?+

which i s shown by Fig. 62.3.

See, f o r example, Y. Sibuya

[ 361.

SUBDOMINAN T SOLUTIONS

266

Fig. 62.3.

Ng.

Fig. 62.3 i s a p a r t o f Fig. 61.5. taining domains.

S2

.

The domain

on F i g . 61.5 .) from

no

Let us denote by

no

62.4.

There a r e f i v e canonical domains con-

no

t h e union of these f i v e canonical

i s shown by Fig. 62.4.

( F i g . 62.4 i s also based

a s u i t a b l e neighborhood of the boundary of

-

no no .

W e c o n s t r u c t a simply connected domain

by removing Then, by u t i -

l i z i n g Theorems 59.1, 60.1 and 60.2, we can f i n d an asymptotic representaI

t i o n of

fZ(x,X) which i s v a l i d uniformly f o r

xE

no

as

X

tends t o

i n f i n i t y i n s t r i p (62.1).

63. Simple t r a n s i t i o n p o i n t problems i n unbounded domains. we constructed two s o l u t i o n s

I n Section 59,

26 7

SIMPLE TRANSITION POINT PROBLEN3

of t h e d i f f e r e n t i a l equation (58.1) i n a s u i t a b l e subdomain i c a l domain and

(59.5)

, where

0

F+(x,A)

F+(x,A) 5 0

,

F (x,X) 5 0

,

{ -

uniformly f o r

X

xE 61 a s

The functions

h(x,h)

F- (x,X)

of a canon-

a r e holomorphic i n (58.17),

tends t o i n f i n i t y i n the s t r i p

pmll 18,

(59.6)

and

61

and

Re[XILpo

9

h(x,-A)

a r e holomorphic and s a t i s f y conditions

(58.19) and (58.19’),r e s p e c t i v e l y , f o r (58.17). ( C f . Theorem 59.1.) Furi s a c o n s i s t e n t canonical domain, we have

thermore, i f

I F+(x,X) - I 141AI(x) I -2N

(60.1) f o r (58.17) and

... , where the 4 are c e r t a i n p o s i t i v e numbers,

N=1,2,

and

=s

-

X

(59.1)

I(x)

p(tI2dt

.

xO

( C f . Theorem 60.1.)

n

On the other hand, i f

i s an i n c o n s i s t e n t canonical

domain, then

N = 1 , 2 , . . ., where A and B a r e two a r b i t r a r y r e a l num4 a r e p o s i t i v e numbers depending on A , while t h e 4 are p o s i t i v e numbers depending on B . ( C f . Theorem 60.2,) The domain &

f o r (58.17) and

bers, and the

i s constructed i n t h e following manner:

...,$ .

2;

j

that ( i ) Lj

(ii) Lj

0 i s mapped

= I(x) onto the whole z-plane c u t by unbounded v e r t i c a l s W e covered each L by a s e c t o r V of c e n t r a l angle 6 s o

conformally by

L1,

?he canonical domain

j

i s contained i n the i n t e r i o r of i s on t h e b i s e c t o r of

(iii) the c e n t r a l angle

6

V

j ’

V

j ’

i s so small t h a t the other

(hf j )

are

SUBDOMINANT SOLUTIONS

26 8

n o t contained i n

V

We assumed t h a t

j .

i s independent o f

6

j

.

Then we

...

a s a simply connected domain d e r i v e d by removing v1 Vk -1 from t h e z-plane, and & = I ( U ) , where x = I-1( a ) i s t h e i n v e r s e of

defined

U

z = I(x;

.

'ihe domain Ip(x)l

(58.10;

p

uhere

i s simply connected.

&I

LB>O

for

X

Furthermore,

,~

E

( C f . S e c t i o n 59.)

i s a c e r t a i n p o s i t i v e number.

In deriving the

asymptotic estimates (59.5), (60.1) and (60.6) a s well as i n c o n s t r u c t i n g t h e twc f u n c t i o n s

h(x,i)

and

, condition

h(x,-A)

(58.10) was indispene-

able.

Therefore, t h e asymptotic r e p r e s e n t a t i o n s (59.4) of t h e two solu-

tions

y+(x,h)

and

y-(x,h)

c f any t r a n s i t i o n p o i n t . that

cannot be used i n an immediate neighborhood

i s on t h e boundary of

xo

n

.

, we

z = I(x)

When we d e f i n e d t h e mapping

assumed

H e r e a f t e r , assuming f u r t h e r t h a t

xo

i s a simple t r a n s i t i o n p o i n t , w e s h a l l c o n s t r u c t asymptotic r e p r e s e n t a t i o n s of

y+(x,k)

borhood of'

and

y-(x,X)

.

xo

which can be used even i n t h e immediate neigh-

As i n S e c t i o n s 58, 59 and 60, we assume t h a t

(63.1)

P(X0)

.

x

but f i x e d polynomial i n

=o

#0

-

xo i s a simple t r a n s i t i o n p o i n t of t h e d i f f e r e n t i a l equa-

This means t h a t

tiori (53.1). A neighborhood of t h e t r a n s i t i o n p o i n t

61.1 .)

n

Let

,

-12

=s

X

(63.2;

z=I(x)

X

p(t) dt 0

onto t h e whole z-plane c u t by unbounded v e r t i c a l s assume t h a t t h e c u t three sectors R

.

i s divided i n t o

61 and 62 by Stokes c u r v e s .

conformally by

in

xo

(Cf. S e c t i o n 61, M g . be a c a n o n i c a l domain f o r e q u a t i o n (58.1) which i s mapped

E0

three ssctors

i s an a r b i t r a r y

We f u r t h e r assume t h a t

P'(X0)

9

p(x)

Lo

, 51

6,

starts from and

z=O

.

Lo,L1,...,4( . We

also

This means t h a t two of t h e

i n t h e neighborhood of

As we d i d i n S e c t i o n 59, l e t us cover

xo

a r e contained

L1, ...,$ by s e c t o r s

. ..

V1, ,Vk of c e n t r a l a n g l e 6 s o t h a t c o n d i t i o n s ( i ) , ( i i ) and (iii)o f S e c t i o n 59 a r e s a t i s f i e d . Furthermore, ue denote by Vo a s e c t o r of cent r a l angle

6

such t h a t

( i f )t h e v e r t e x of (ii') (iii1 )

the cut

Lo

Vo

is at the point

bisects the sector

the c e n t r a l angle

tained i n

Vo

.

6

Vo

z=O

,

,

is so small that

L1,

...,\

a r e n o t con-

We s h a l l then d e f i n e a simply connected domain

Uo by

A METHOD DUE TO T.M.

removing

Vo"Vlw.

kV'.

n

(63.3)

Bo=I-L(uo)

.

8,

Finally, we

by

Our main concern i s the asymptotic behavior o f uniformly f o r

269

(See Fig. 63.1 .)

from z-plane. of

define a subdomain

CHERRY

y+(x,A)

and

y- ( x , a )

as 1 tends t o i n f i n i t y i n s t r i p (59.6).

x E Bo

LD

b

Fig. 63.1. I n Section 66, we s h a l l a l s o i n v e s t i g a t e t h e case when the v e r t i c a l cut

Lo

.

This

contains only one o f the three s e c t o r s 60

, 5

, but

means t h a t t h e domain

n

z = O i s not t h e s t a r t i n g point of

and 62 i n the neighborhood of the t r a n s i t i o n point

64.

i s on

z=O

A method due t o T.M. Cherrx.

xo

Lo

.

I n Section 58, w e constructed the two

formal s o l u t i o n s (58.4) and (58.5) o f the R i c c a t i equation (58.3). construction, we used the condition:

lp(x)I 2 p> 0

for

xE

&

.

I n this

Since the

is on t h e boundary of t h e domain Po , w e can not use this method i n go Instead we s h a l l follow t h e scheme due t o T.M. Cherry.* To begin with, we consider a d i f f e r e n t i a l equation transition point

(64.1)

xo

.

2 d w/d5'-

Changing the v a r i a b l e

*

sw= O

.

5 by

See A . P d 6 l y i , M. Kennedy and J . L . McGregor [lo; $4. pp. 465-471], W. Wasow [47; Chapter V I I I , pp. 157-1941, and T.M. Cherry [6].

SUBDOMINAN T SOLUTIONS

270

we d e r i v e from (64.1) t h e d i f f e r e n t i a l e q u a t i o n

Eu'ote t h a t

(64.4) Next, changing t h e unknown q u a n t i t y

w

by

--

1

(54.5)

w=J(x)

'h ,

we d e r i v e from ( 6 44 ) t h e d i f f e r e n t i a l e q u a t i o n

(64.6) On t h e o t h e r hand, i f we change t h e unknown q u a n t i t y

y

by

_(64.7)

Y=P(x)

4v

,

the d i f f e r e n t i a l equation

(58.1)

y"

- i2p( x ) y = 0

is reduced t o

-

This equation can be w r i t t e n a s

(64.8)

We s h a l l transform t h e d i f f e r e n t i a l e q u a t i o n (64.8) i n t o an i n t e g r a l equation i n the next s e c t i o n . Let us prove t h e f o l l o w i n g lemmas.

LEMMA 6C.1:

The f u n c t i o n

(64.9) i s holomorphic i n a neighborhood of t h e t r a n s i t i o n p o i n t x = x 0 Proof:

Put a=pl(xo)

,

b=pll(xo)

Assumption ( 6 3 .l) i m p l i e s t h a t

.

.

A METHOD DUE

TO T.M. CHERRY

271

.

afo Since

1 2 p(x) = a(x-xo) + zb(x-xo) i n t h e neighborhood o f

-1

, 11.

+ O[ (x-x,)

3

]

x=xo

1

2 1-2 2 ~[ a) +-a 4 b(x-xo)+O[(x-xo) 11

2 ~ ( X ) ~ = ( X - X

,

and

1 I ( x ) = ( x - x o ) 3 ~ 2 p a ~ + ~ ( x -+x o[ o )(x-x,) 2 31 3 1oa2

-

.

Hence b 0

Thus, 5a x-xo i n the neighborhood of

x=xo

.

+ o(1)

On t h e o t h e r hand,

0

and

i n t h e neighborhood o f

i s holomorphic a t

x=x

x=xo

.

0

.

The n o t a t i o n

O(1) i n d i c a t e s a term which

Since

we g e t q(x) = O(1) i n t h e neighborhood of

6~.2:

If

in

9,

Proof:

, where

.

This proves Lemma 64.1.

R i s a s u f f i c i e n t l y large p o s i t i v e number, w e have

I$$

(64.10)

x=xo

K

IKJI(X)I-~

for

I I W J2~

i s a c e r t a i n p o s i t i v e number.

It is easy t o s e e t h a t q(x)

= o(x-2)

SUBDOMIN AN T SOLU TIONS

272

as

x

tends t o i n f i n i t y i n

Bo

.

as

x

tends t o i n f i n i t y i n

&lo

, where

respect to

x

.

Hence

n: i s t h e degree o f

p(x)

with

On t h e o t h e r hand,

I(X) = o ( x as

tends t o i n f i n i t y i n

x

If

LEMMA 6L.3:

to

This proves Lemma 64.2.

ro is a s u f f i c i e n t l y small p o s i t i v e number, w e have

for


, where H

Proof:

.

I$$

(64.11)

&

Bo

11(x)l < r o

i s a c e r t a i n p o s i t i v e number.

By v i r t u e o f lemma 64.1, t h e f u n c t i o n

neighborhood o f

x =x

i s holomorphic i n t h e

Hence

0 .

i n t h e neighborhood o f

q(x)

x = x0

.

On t h e o t h e r hand,

I(x) =0[(x-xo) 3/21 This proves Lemma 6 4 . 3 .

65.

Assmptotic s o l u t i o n s i n t h e domain

a0

.

Since t h e domain

Uo does

Vo , w e can f i n d an i n t e g e r k such t h a t 1 1 [: ( 2 k - l ) n t 6 1 < a r g [ I ( x ) ] < (Z(kt1)t l)n - 6 1

not contain the s e c t o r

z[

(65.1) for

.

xE Lo

i s odd.

wards i f

k

of

in

p(x)

The c u t

.

Lo

i s d i r e c t e d downwards i f

d3

i s even. and u p

is determined by t h e g l o b a l behavior

In t h e l a s t s e c t i o n , w e defined

J ( x ) = [2. * I ( x )3

(65.2)

k

The i n t e g e r

k J(X)

.

Hence, (65 .l) i m p l i e s t h a t

(65.3) for x E b o

.

This means t h a t

-( n - 46) <

-?<

a r g [ J( x) ] -

2kn 1 y< n - -6 , 3

(65 .4)

< a r g [ J ( x ) 1 - 2(k+l)n < for x E a O

.

3

- 716

by

ASYMPTOTIC S O L U T I O N S I N THE DOMAIN

b

I n Section 7, we defined the solutions

m,k

27 3

go

(x,a)

of the d i f f e r e n t i a l

equation y"-P(x)y=O

m-1

,

~ ( x =) x m + y x

Let u s consider t h e s o l u t i o n s Id

({,al) 1,k d2w/ds2- (st al)w=O

of the d i f f e r e n t i a l equation

.

(65.5)

By v i r t u e of Theorem 7.1, t h e s o l u t i o n conditions :

b 1 ,k (S,al)

( i ) l ~ ~ , ~ ( S , ais , ) an e n t i r e function of

b,,,(!

(ii)

+ ...+am .

,al)

and

b

1,k

( 5 ,al)

s a t i s f i e s t h e following

(s,al) ;

admit asymptotic representations

uniformly on each compact s e t i n a plane a s 1sector

x

tends t o i n f i n i t y i n the

where uo = exp[ i

(65.8)

2

~

]

and (65.9) ( C f . Section 21, Lemma 21.1.)

I n Section 8, w e proved t h a t

b1 ( 5 , all = 2 Lfi( S t y )

(8.4) where

Ai(x)

i s A i r y ' s i n t e g r a l given by (8.1). Since

we have

( C f . (8.5) i n Section 8.)

We s h a l l now prove the following lemmas

-65.1: The d i f f e r e n t i a l equation (64.6) has two l i n e a r l y independent solutions

SUBDOMINANT SOLUTIONS

1

u = \(x,h)

1

= A ?r J ( ~ ) ~ n i ( w ~ \ ~ / ~ J, ( x ) )

(65.11)

1

1

c

.

u = ~ k + ~ ( x =, lh ) J ( X ) ~ A ~ ( L I - ~ - ~ X ~ / ~

J(x))

0

The d i f f e r e n t i a l equation (64.6) was derived from (64.3) by trans-

Proof:

formation (64.5). If we s e t a1 = 0 and { =A2/3J(x) a s o l u t i o n of (64.3). This proves Lemma 65.1. LEMMA 65.2:

Assume t h a t (65.1) i s s a t i s f i e d i n

(65.12)

O < 60/~0(

tan(z6) 1

do

i n (65 .lo), we g e t

, and

that

.

i n the domain (65.13) we have

(65.14)

and -

(65.15)

gg

AI(x) tends t o i n f i n i t x .

Proof:

Since (65.1) i m p l i e s (65.4), we have I ~ r g [ h ~ ' ~ J ( x ) ]
{

= hI(x) , 1 ($)1/3[~2/3~(x) = [AI(x) 11/3

5[A2/'J(x)

13/'

32

,

Since

ASYMPTOTIC SOLUTIONS I N THE DOMAIN

275

Po

and

1

1 1

,

[ X 2/3 J ( x ) ]4 = X 6 J ( x ) ~

\(x,h)

t h e asymptotic representation of

can be derived immediately from

.

t h e asymptotic representation of 1, (5,O) ( C f . (65.6) .) The a s p p t o 19k t i c r e p r e s e n t a t i o n of %(x,X) i s obtained by d i f f e r e n t i a t i n g t h a t of % with r e s p e c t to x

.

( C f . Theorem

i s similar t o t h a t of ( 6 5 . u ) .

3.3, Chapter 1.) The proof o f (65.15)

This completes the proof of Lemma 65.2.

we use more accurate information of

--

If

Airy's i n t e g r a l such a s (8.3) i n Sec-

1

I

t i o n 8, we can reduce O([XI(x)] ') t o O([xI(x) ]-I) L e t us now compute the Wronskian of % and \+l

.

.

Since these two

functions a r e s o l u t i o n s of ( 6 4 . 6 ) , the q u a n t i t y

--2

dX) [Uk%tl- %\+lI must be independent of

as

AI(x)

x

.

Furthermore,

1 7-k

tends t o i n f i n i t y , where we used the i d e n t i t y

-3

k (-1) = w o

,

Hence

Let us prove t h e following theorem.

'JHEOREM 65.1;

Assume t h a t

(65.1) is s a t i s f i e d i n Bo

, and

that

po

a

Then the d i f f e r e n t i a l eauation

s u f f i c i e n t l y large p o s i t i v e number.

2 yl1-h p ( x ) y = O

(58.1)

has two s o l u t i o n s of t h e forms

1

f

1

y=yk(x,lr) = X p ( x )-4J(x)4[1+Gk(x,X) lni(w~kh2'3J(x))

where t h e functions the conditions

z

1

Gk(x,A)

'& %tl(x,X)

a r e holomorphic and s a t i s f y

SUBDCMINANT SOLUTIONS

276

in domain (65.13), and the q u a n t i t y c i s a c e r t a i n Dositive constant.

Proof:

is odd. First o f all, we reduce (58.1) t o (64.8) by (64.7). Then we reduce (64.8) t o a n We s h a l l construct

yk(x,k)

assuming t h a t

k

i n t e g r a l equation

k -71

v(x,h)

and this i n t e g r a l equation is reduced t o k--1 2nw 4

(65.20)

cp(x,h) =

1 +x 0 J

- cp(t,A)dt

C(x,t,a)

Y(X)

PW2 by the transformation

(65.21) srhere

q(x)

.(.,A)

=~(x,L)cp(x,A) ,

i s given by (64.9), and

(65.22) We choose the path of i n t e g r a t i o n y ( x ) We s h a l l estimate the function

(65.23) Since tion

Ai(5) \!x,X)

ciently large.

xEBo

,

tE8,

G(x,t,A)

,

i n 'the same way as i n Section 59. for

(Im[Al(lBo

,

Re[AI1p0

.

has zeros on the negative r e a l a x i s i n <-plane,* the funcdoes not have any zero i n domain (65.13) if

po

i s suffi-

Note t h a t i f (65.1) and (65.12) are s a t i s f i e d , then (65.16)

holds i n domain (65.13). Hence, by using Lemma 65.2, we g e t

i n domain (65.13), where

c1

i s a c e r t a i n p o s i t i v e constant.

manner, we obtain

*

See, f o r example, M. Abramowitz and I.A. S t e w [l; P. 4501.

In the same

277

ASYMPTOTIC SOLUTIONS IN THE DOMAIN Bo

l u . , p , x ) = d - a I ( x ) 11

(65 2 5 )

5c2

and

1%

(65.26)

+1(x,a

)~.&x)l I c3

i n domain (65.13), where

c2 and

c

are c e r t a i n p o s i t i v e constants. I n 3 deriving (65.2,$), (65.25) and (65.26), w e used the assumption t h a t k i s odd. Furthermore, i n deriving (65.25) and (65 .%), we regarded the two

] and %+1 (x,h)%(x,~) as functions of Let us change the v a r i a b l e t i n (65.20) by

quantities

hI(x)

.

%(x,h)exp[-XI(x)

c =I(t)

(65 27)

.

Then t h e i n t e g r a l equation (65.20) becomes

.

z = I(x) and C( z) = I ( y ( x ) ) I n Section 59, we explained how t o choose C(z) Along C(z) , the i n e q u a l i t y Iexp[-M(z<)]l 51 holds, s i n c e Re[AC] is s t r i c t l y decreasing along C(z) This means t h a t G(x,t,h) is bounded by a constant f o r where

.

.

,

xEao

,

tCy(x)

IImbIl

56,

R~[A]LP,

.

Also, by v i r t u e of Lemmas 64.2 and 64.3, we g e t

(65 29) where

z = I(x) and c

4

i s a c e r t a i n p o s i t i v e constant.

construct a bounded s o l u t i o n cp(x,k)

is s u f f i c i e n t l y large.

Therefore, we can

of (65.20) i n domain (65.13), i f

po

Then (65.20) again implles

cp(x,x) = 1+ o(x-l) uniformly f o r

x E B o as b

I M X I 1 56, Hence

tends t o i n f i n i t y i n the s t r i p 9

R e [ x I L P,

--1

Y = Y k ( x , x ) =p(x) 4qx,X)'p(x,a)

s a t i s f i e s all requirements given i n Theorem 65.1.

*

Cf. Lemma 59.1.

Similarly, we can t r e a t

278

SUBDOMINANT SOLUTIONS

.

Yk f o r an even k as w e l l a s yk+l

This completes t h e proof of Theorem

65.1. I n the same way as we proved Theorems 60.1 and 60.2,

we can prove the

following two theorems.

If n

65.2:

(65.30)

f o r (65.13), =ORPI

I

65.3:

f o r (65.13),

i s a c o n s i s t e n t canonical domain,

where

Jf

where

L E IAIWI-’

(X,OI

E

i s a c e r t a i n p o s i t i v e number.

n

i s an i n c o n s i s t e n t canonical domain,

B

A

p o s i t i v e number dependina on on -

,

l%(x,A)I G I A I ( x ) l - l

‘k+l

then

are

A

?

then

two a r b i t r a r y real numbers and

while 6’

i s a p o s i t i v e number dewndine,

.

B

As w e s t a t e d a t t h e beginning of S e c t i o n 63, t h e d i f f e r e n t i a l equation

(58.1) has the two s o l u t i o n s ( 5 9 . 4 ) which s a t i s f y conditions (59.5) and (60.1) o r conditions (59.5) and (60.6) i n t h e domain (58.17)

where

a

xE

9

II d h l l l 6 ,

9

Re[X12Po

.

W e a l s o explained how t h e domain It i s easy t o see t h a t B i s contained i n a0 Let

P is a s u i t a b l e subdomain o f Cl

i s constructed.

9

.

us consider the s o l u t i o n

where the path of i n t e g r a t i o n f o r ity in

h(x,A)

i s a curve j o i n i n g

a so that X

l i m s h(t,h)dt=O

as Re[I(x)]

tends t o

(P

x

to infin-

-.

Then, by v i r t u e o f the uniqueness of subdaminant s o l u t i o n s , we g e t

r

0

if

k

is odd,

FaMARKs

279

By using these formulas, we can find t h e asymptotic behavior o f y+(x,X) a s a tends t o i n f i n i t y i n domain (65.13). Similarly, we can f i n d the

-

y (x,a)

asymptotic behavior of

tends t o i n f i n i t y i n domain (65.13).

as

66. Remarks on the case when n contains only one of the t h r e e s e c t o r s 60 , E j E2 i n t h e neighborhood of the s i m D l e t r a n s i t i o n p o i n t xo I n s e c t i o n s 63, 64 and 65, we assumed t h a t t h e v e r t i c a l c u t Lo starts Note t h a t z = O i s t h e image of t h e t r a n s i t i o n p o i n t xo by from z = O the mapping z 2 I(x) I n o t h e r words, I (x,) = 0 As we mentioned before,

.

.

.

.

n

this assumption means t h a t

contains two of the t h r e e s e c t o r s

and E2 i n the neighborhood of the simple t r a n s i t i o n p o i n t s e c t i o n , we s h a l l b r i e f l y discuss the case when these t h r e e s e c t o r s i n t h e neighborhood of

xo

xo

.

e0 , I n this

contains only one of

.

For example, i f

n

is

one of t h e canonical domains given by Figures 61.7 and 61.8, t h e r e are four simple t r a n s i t i o n p o i n t s

n

The domain

,

x2

x3

and

x

W e assume t h a t the canonical domain mapping

(66.1)

z=I(x)

n

xl

hood of

, and

t h e domain

n

0

i s on t h e c u t Lo

i s a non-zero real number.

if T ~ 0>

4'

i s mapped conformally by the

onto t h e whole z-plane c u t by unbounded v e r t i c a l s T~

x

=l p ( t ) 2 d t X

z=O

and

-1

X

where

.

contains two of those t h r e e s e c t o r s i n the neighborhood of

x

assume t h a t

0

on the boundary of

4

However, i n the neighborhood of 3 contains only one of those t h r e e s e c t o r s .

x2

and

,

x1

upwards i f

xo i n the domain

T ~ 0<

n

.

.

, and

that

The c u t Let

Lo,%, Lo

...,4, . We

starts from

also

z=iTO

,

Lo i s d i r e c t e d downwards

B(xo) denote a small neighbor-

If this neighborhood

8(x0) is s u f f i -

c i e n t l y small, we have e i t h e r R e [ I ( x ) ] >0 f o r every x E & ( x o )

or

Re[I(x)] < 0 f o r every x E B(xo)

(66.2)

We assume t h a t ( 6 6 . 2 ) is s a t i s f i e d . domain

6,

of

s e c t o r s V1,. and 63 .) such t h a t

Let

We s h a l l construct a s u i t a b l e sub-

0 i n the follovfng manner.

..,Vk

.

W e s h a l l cover

i n the same way as we d i d before.

z =iT1

5 ,...,4,by

( C f . Sections

59

be a fixed p o i n t on t h e imaginary a x i s i n z-plane

280

SUBDOMINANT SOLUTIONS

(66.3) Denote by

B1

the semi-disc defined by

(66.4)


]z-iTol

r

where

.

iT1$Lo

0

i s a small p o s i t i v e number s a t i s f y i n g t h e condition 1

0 < ro < 7 d n ( l

(66.5) ?he point

,

iT1

Assume t h a t

1 'T1 - To) 3

Tol

a

d i v i d e s the imaginary axis i n t o two p a r t s

LOCI1

.

I1 and I2

.

B2 a s e c t o r s a t i s f y i n g t h e following

Denote by

condi ti ons :

B2

i s a t the p o i n t

(a)

the vertex of

(b)

the c e n t r a l angle of

(c)

IL is on the boundary of

(d)

Re[z]>_O f o r

Also, fienote by

(a0 (b')

B

~

3

the vertex o f

fi2

€

.9

z=irl

~

a s e c t o r s a t i s f y i n g the following conditions:

B

is a t t h e p o i n t z = O 1 , a3 i s 56

3 the c e n t r a l angle of

( c ' ) one of the two s i d e s of t h e s e c t o r ( d ~ )Re[z]
Set

€

.8

vo = 81" a2"

B3

for

~

-

(66.6)

,

1 , 56 B2 ,

is

, i s contained i n

B3

Lo

,

~

.

(See F i g . 66.1.)

0

Fig. 66.1. W e s h a l l d e f i n e a simply connected domain from z-plane

.

'Ihe domain

io

i s given by

0

by removing

..

Tou V1u . u Vk

EXAMPLES

281

-s 0 = 1-1(uo) - .

(66.7)

I t i s easy t o see t h a t we can j o i n each p o i n t

in

z

Go

to

to

in

m

by

a curve

,

c=c(s)

so t h a t

c(o)=z

,

o<s<scD

However, i t i s not possible t o

Re[aC(s)] i s s t r i c t l y decreasing.

j o i n each increasing.

io t o

in

z

in

m

to

so that

Re[AC(s)]

E0

This i s the d i f f e r e n c e between

Section 63 ( o r the domain

U

of Section 59).

construct t h e two s o l u t i o n s yk

and

is strictly

and the domain

of

Uo

Due t o this f a c t , we cannot

of (58.1) i n the domain

yktl

(66.8) However, w e can construct one of these s o l u t i o n s . t i o n (65.1) is s a t i s f i e d f o r

xE

zo , and

i n the same way as i n Section 65.

If

destroyed the symmetry of t h e domain.

is odd, w e can construct

i s even, we can construct

k

Note t h a t w e removed

i n domain (66.8).

k

For example, i f condi-

81

from z-plane.

W e had t o remove

Thus we

9 , since

i s the image of another t r a n s i t i o n point on t h e boundary of

n

67.

Re[ I ( x ) ] > 0 f o r

Examples.

x E S(xo)

The case

can be t r e a t e d i n a s i m i l a r manner.

As i n Section 62, w e s h a l l i l l u s t r a t e how t o use the method

of Sections 63, 64, 65 and 66 by using case (62.6). asymptotic representation of

st" S-

x1

and

S+"

- , we

S

, we

constructed the domain

x2 on the boundary of

no

by removing from

the domain

fo(x,X)

I n order t o derive an

no

of

Flg. 62.1 c l e a r l y shows t h a t there are two t r a n s i -

(62.8) i n Section 62. tion points

iTo

.

So far w e have assumed t h a t condition (66.2) i s s a t i s f i e d . when

yk

yktl

n

.

W e derived t h e domain

a s u i t a b l e nsighborhood of t h e boundary.

In

could u t i l i z e the method of Sections 59 and 60.

Due t o the reason s t a t e d i n Section 66, t h e method of Sections 63, 64, 65

and 66

does not apply t o

fo(x,X)

i n the neighborhood of

x1 , w e can apply the method t o which i s shown roughly by Fig. 67.1. transition point

fo(x,X)

x2

.

As t o the

i n a domain

SUBDOMINANT SOLUTIONS

282

I

\

Fig. 67.1.

By using the same method, w e can d e r i v e some formulas for the parab o l i c cylinder function

D

(67.1)

Dv(x)

such a s

-1

1(2v 2x) V-2

where

The branches of t i v e for x > 1

2

.

(x

given by F i g . 67.2 uith

x =1 a d

1

-1

- 1)2 and

2

(x

-1

- 1)4

are chosen so t h a t they a r e posi-

Fonmiia (67.1) i s v a l i d uniformly for

as

tends t o

v

x =-1 2

.

Sm

.

Mg. 67.2.

x

i n a domain

Fig. 67.2 i s based on Fig. 61.3

283

EXAMPLES Formula (67.1) i s one of the r e s u l t s obtained by A . Erdhlyi, M. Kennedy and J . L . McGregor."

The method of Sections 63, 64, and 65 i s a

straight-forward g e n e r a l i z a t i o n of t h e i r method.

.

This s o l u t i o n i s subdominant f2(x,A) S2 given by Fig. 62.3. As w e mentioned i n Section 62,

L e t us consider the s o l u t i o n

i n the domain

t h e r e a r e f i v e canonical domains containing canonical domains i s the domain

no

S2

.

The union of these f i v e

shown by Fig. 62.4.

t r a n s i t i o n p o i n t s on the boundary of

no

.

of Sections 63, 64, 65 and 66 a t a t r a n s i t i o n p o i n t canonical domains which contain I n case of

x

3

(or x

4

, and

S2

, we

J

no

.

as

A

.

I n case of

x2

tends t o i n f i n i t y

x i n a domain derived from

by removing a s u i t a b l e neighborhood of t h e boundary of

( o r x4 )

x

By u t i l i z i n g the method, we

f2(x,A)

i n s t r i p (62.1) which i s v a l i d uniformly f o r

**

consider a l l

whose boundaries contain

can f i n d an asymptotic representation of

x3

x

j ' ) , all of the f i v e canonical domains s a t i s f y the

requirements, and hence t h e i r union i s

no

There a r e s i x

I n order t o u t i l i z e the method

no

except a t

(or x5 ), t h e r e are two canonical

domains which s a t i s f y the requirements.

.

( o r n: ) The domain n : i s given by Fig. e can f i n d an asymptotic i s given by Mg. 67.4. W 67.3, and t h e domain r e p r e s e n t a t i o n of f 2 ( x , i ) which i s v a l i d uniformly f o r x i n a domain domains i s denoted by

derived from

n; n;

(or

n;

The union of these two canonical

n: n:

) by removing a s u i t a b l e neighborhood o f the

x ( o r a t x ) . Due t o the reason 2 5 s t a t e d i n Section 66, the method does not apply t o f2(x,A) a t x1 and

boundary of

(or

) except a t

Fyg. 67.3.

*

C f . A . Ekde'lyi, M. Kennedy and J . L . McGregor [lo; (8.8), p.477, and Flg. of t h e domain 8, f o r arg v = 0 on p.4821. ** This means t h a t the representation i s v a l i d i n the neighborhood of x3

.

SUBDOMINANT SOLUTIONS

284

//

fl 0

f -T 8 4

F i g . 67.4.

Figures 67.3 and 67.4 are based on F i g . 6 1 . 5 .

1

3

3 7

REFEFENCES This l i s t contains only works t h a t were used by the author i n the preparation of this book. More complete references on t h e asymptotic solut i o n of ordinary d i f f e r e n t i a l equations a r e found, f o r example, i n such books as F.W.J. Olver [31] and W. Wasow [47].

1.

M. Abramouitz and I . A . Stegun ( E d i t o r s ) , Handbook.of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Applied Mathematics Series 55, Third P r i n t i n g , March 1965, with corrections; U S . Government P r i n t i n g Office, Washington, D C

..

2.

I . Bakken, A multiparameter eigenvalue problem i n the complex plane, D i s s e r t a t i o n and Technical Report, University o f Minnesota, 1974.

3.

L. Bieberbach, Theorie der Gew%nlichen Springer-Verlag, 1953.

4.

G .D

5.

G.D. Birkhoff, The generalized Riemann problem f o r l i n e a r d i f f e r e n t i a l equations and the a l l i e d problems f o r l i n e a r d i f f e r e n c e and q-difference equations, Proc. Amer. Acad. Arts and S c i . , 49(1913) 521-568.

6.

T.M. Cherry, Uniform asymptotic formulae f o r functions with t r a n s i t i o n p o i n t s , Trans. h e r . Math. SOC., 68(1950) 224-257.

7.

E.A. Coddington and N . Levinson, Theory of Ordinary D i f f e r e n t i a l Equations, McGraw-Hill, 1955.

8.

G . E l M n g , h e r e i n e Klasse von Riemannschen FlPchen und i h r e Uniformisierung, Acta SOC. S c i . fenn. (N.S.) 2, No. 3(1934) 5-60.

9.

A. Erde'lyi, Asymptotic ExDansions, Dover, New York, 1956.

10.

A . Erdklyi, M. Kennedy, and J . L . McGregor, Parabolic cylinder f'unct i o n s of l a r g e order, Jour. Rat. Mech. Anal., 3(1954) 459-485.

Differentialgleichunpen,

.

Birkhoff, Singular points of ordinary l i n e a r d i f f e r e n t i a l equat i o n s , Trans. Amer. Math. SOC., l O ( l 9 O q ) 436-470.

11. M.A. Evgrafov and M.V. Fedoryuk, Asymptotic behaviour a s A + of the s o l u t i o n of t h e equation w"( z) - p(z,X)w(z) = 0 i n the complex z-plane, Russian Math. Surveys, 21(1966), 1-48, t r a n s l a t e d from Uspehi Mat. Nauk., 21(1966), No. 1 (127) 3-50. 12.

K.O. Friedricks, Special Topics i n Analysis, P a r t B, Asymptotic Intep a t i o n of D i f f e r e n t i a l Esuations, New York University, 1953-54.

13. H.E. Gollwitzer and Y. Sibuya, Stokes m u l t i p l i e r s f o r subdominant s o l u t i o n s of second order d i f f e r e n t i a l equations with polynomial c o e f f i c i e n t s , Jour. fcr d i e r e i n . und ange. Math., 243(1970) 98-119.

REFERENCES

286

u.P . 15.

Hartman, Ordinary D i f f e r e n t i a l Equations, John Wiley, 1%4.

J . Heading, An Introduction t o Phase-Internal Methods, Methuen, London, 1962.

16. E . H i l l e , Lectures on Ordinary D i f f e r e n t i a l Equations, Addison-Wesley, 1969. 17.

P.F. Hsieh and Y. Sibuya, On the asymptotic i n t e g r a t i o n of second order l i n e a r ordinary d i f f e r e n t i a l equations with polynomial c o e f f i c i e n t s , J . Math. h a . Appl., 16(1966) 8.4-103.

18. P.F. Hsieh and Y. Sibuya, Note on r e g u l a r perturbations of l i n e a r ordinary d i f f e r e n t i a l equations a t i r r e g u l a r s i n g u l a r p o i n t s , Funkcia1 Ekvac , 8 ( 1966) 9-108.

.

19.

P.F. Hsieh and Y. Sibuya, On d i s t r i b u t i o n of zeros of s o l u t i o n s of a second order l i n e a r d i f f e r e n t i a l equation with polynomial c o e f f i c i e n t s , J . Math. Ana Appl., 43( 1973) 101-113.

.

20.

J.A. Jenkins, Univalent Functions and Conformal Mapping, Second P r i n t i n g Corrected, Springer-Verlag, 1 % 5 .

21.

T. Kimura, On the i t e r a t i o n of a n a l y t i c functions, Funkcial Ekvac., l4( 1971) 197-238.

22.

R .E. Langer, The asymptotic s o l u t i o n s of ordinary l i n e a r d i f f e r e n t i a l equations of the second order, with s p e c i a l reference t o a turning p o i n t , Trans. Amer. Math. SOC., 67(1949) 461-490.

23.

R.W. McKelvey, ?he s o l u t i o n s of second order l i n e a r ordinary d i f f e r e n t i a l equations about a turning p o i n t of order two, Trans. Amer. Math.

SOC., 79(1955) 103-123.

24. A . I . Markushevich, Theory of Functions of a Complex Variable, Vol. 11, Translated from t h e Russian by R.A. Silverman, Prentice-Hall, 1965. 25.

F.E. h l l i n , On t h e r e g u l a r perturbation of the subdominant s o l u t i o n t o second order l i n e a r ordinary d i f f e r e n t i a l equations with polynomial c o e f f i c i e n t s , F’unkcial Ekvac., 11(1968) 1-38.

.

26.

R . Narasimhan, Several Complex Variables, The University of Chicago Press, 197l.

27.

R . Nevanlinna, h e r Riemannsche Flachen mit Endlich Vielen Windungspunkten, Acta Math. 58(1932) 295-373.

28.

R . Nevanlinna, Analytic Functions, Translated by P . W g , SpringerVerlag , 1970.

29.

T. Nishino, Nouvelles recherches sur les fonctions e n t i b e s de plus i e u r s v a r i a b l e s complexes ( I ) , J. Math. Kyoto Univ., 8(1%8) 49-100.

30. F.W.J. Olver, Error a n a l y s i s of phase-integral methods. I . General theory f o r simple turning p o i n t s , J o u r . of Res. of the Nat. Bureau of Stand., B, 69B(1%5)271-290.

REFERENCES

287

31.

F.W.J. Olver, Asymptotics and Special Functions, Academic P r e s s , 1974.

32.

Y. Sibuya, Second order l i n e a r ordinary d i f f e r e n t i a l equations cont a i n i n g a l a r g e parameter, Proc Japan. Acad., 34(1958) 229-234.

33.

Y. Sibuya, Simplification of a system of l i n e a r ordinary d i f f e r e n t i a l equations about a s i n g u l a r p o i n t , F u n k c i d . Ekvac., 4(1962) 29-56.

34.

Y. Sibuya, Subdominant s o l u t i o n s of l i n e a r d i f f e r e n t i a l equations with polynomial c o e f f i c i e n t s , Mich. Math. Jour., ( +!,.I 1967) 53-63.

35.

Y. Sibuya, Stokes m u l t i p l i e r s of subdominant s o l u t i o n s o f t h e d i f f e r e n t i a l equation y t t - ( x3 t A)y=O , Proc. Amer. Math. SOC., 18(1%7) 238-243

36.

Y. Sibuya, Subdominant s o l u t i o n s of the d i f f e r e n t i a l equation 2 .(x-am)y = 0 , Acta Math., 119(1967) 235-272. y" - A (x-a,) (x-a,)

37.

Y. Sibuya, Perturbation of l i n e a r ordinary d i f f e r e n t i a l equations a t i r r e g u l a r singular p o i n t s , Funkcia1 Ekvac , 11(1968) 235-246.

38.

Y. Sibuya, A boundary value problem i n t h e complex plane, Proc. of Conf. a t Western Michigan Univ., Kalamazoo, April 30-May 2, 1970, Lecture Notes i n Math., No. 183, 128-144, Springer-Verlag, 1971.

39.

Y. Sibuya, Perturbation a t an i r r e g u l a r s i n g u l a r p o i n t , Proc. of Japan-U.S. Seminar on Ordinary D i f f e r e n t i a l and Functional Equations, Kyoto, September 6-11, 1 9 7 l , Lecture Notes i n Math., No. 243, 148168, Springer-Verlag, 1971.

40.

Y. Sibuya, Uniform s i m p l i f i c a t i o n i n a f u l l neighborhood of a t r a n s i t i o n p o i n t , Amer. Math. SOC., Memoir l.49, 1974.

41.

Y. Sibuya, Subdominant s o l u t i o n s admitting a prescribed Stokes phenomenon, Proc. of I n t e r n a t i o n a l Conf. on D i f f e r e n t i a l Equations a t Univ. of Southern California, Sept. 3-7, 1974, 709-738, Academic Press, 1975.

42.

Y. Sibuya and R.H.

.

..

.

Cameron, An e n t i r e s o l u t i o n of the f b c t i o n a l equa-

-

t i o n f ( X ) + f ( w X ) f ( w 'a) = 1 , ( w 5 =I) , Proc. of symposium on ~ r d i nary D i f f e r e n t i a l Equations a t Univ. of Minnesota, May 29-30, 1972, Lecture Notes i n Math., No. 312, 194-202, Springer-Verlag, 1973. 43

L. Sirovich, Techniaues of Asymptotic Analysis, Applied Math. S c i . 2, Springer-Verlag, 197l.

44.

C.A. Swanson and V.B. Headley, A n extension of Airy's equation, SIAM J. Appl. Math., 15(1967) l4OO-Ul2.

45 ' E.C. ntchmarsh, Haenfunction E b a n s i o n s Associated with Second-

Order D i f f e r e n t i a l Eauations, P a r t I, Second Edition, Oxford, 1962.

46.

.

F .G Tricomi, Fonctions Hypera6om6triaues Confluentes , Mem. des Sci Math., No. 140, Gauthier-Villars, 1960.

.

288

FEEFENCES

47.

W . Wasow, AsymDtotic E h a n s i o n s f o r Ordinary D i f f e r e n t i a l Ehuations, John Wiley, 1965.

48.

W. Wasow, The asymptotic t h e o r y of l i n e a r d i f f e r e n t i a l e q u a t i o n s i n v o l v i n g a parameter, Le Mat., ll(1955) 134-148.

49.

W. Wasow, Simple turning-point problems i n unbounded domains, SIAM J . Math. Anal., l ( 1 9 7 0 ) 153-170.

50.

F . Zernike, Eine asymptotische Entwicklung fi d i e g r a s s t e N u l l s t e l l e 3.er Hermiteschen Polynome, Koninklijke Akademie van Wetenschappen t e Amsterdam, Proc. of t h e S e c t i o n of S c i . , 34(1931) 673-680.

Airy's integral

22( d e f i n i t i o n ) , 53, 75( Theorem 19.1, ( i v ) ) , 87( connection formula, Theorem 22.1), 168, 203, 273(Lemma 65 -1), 274 (Lemma 65.2), 275( Theorem 65.1)

Asymptotic expansions

1, 12, 14

Asymptotic s o l u t i o n s a t an irregular singular point

&(Theorem 11.1), 228( Theorem 56.1), 229 ( Theorem 56.2)

Asymptotic values of a meromorphic function

173(Remark 1)

Birkhoff 1s problem

231( Problem 56.1)

Boundary value problems

45, 46, 128, 167(Problem 39.1)

Canonical domains

243(Definition 59.3), 249(consistent o r i n c o n s i s t e n t , Definition 60 .l)

Cherry's method

269

Dynamical systems

254, 256

Formal power s e r i e s

3, 10

Hankel functions

94

Index of a closed curve

256

Inverse of an a n a l y t i c mapping

198

I v e r s e n l s property

156

Koebe 1s d i s t o r t i o n theorem

192

Koebe 1s theorem

194

Line complex (topological tree)

186

Logarithmic ends

179

Multipaxameter eigenvalue problem Nevanlinnals problem

197 174(Problem 40 .l)

Nevanlinna1s theorem

175( Theorem 40.1)

Parabolic cylinder functions

2,$(definition), 53, 75(Theorem 19.1, ( i v ) ) , m( connection formulas, Theorem 22.2), 97, 129, 169, 203, 282(fomula (67 .l) due t o Erdhlyi-Kennedy-McGregor)

Phragme'n-Lindel'df theorem H c a r d 1s exceptional values

150 171, 173(Remark 2)

Poincar6-Bendixson theorem

256

Nadratic differentials R i c c a t i equation

254 28(Lemma 9.1), 236(Lemma 58.1), 241 (Lemmas 58.2 and 58.3)

Rouch6 1 s theorem

133

290

INDEX

S i n g u l a r i t y of a multiple-valued function

153

Schvarzian

172

Stokes curves

242 ( D e f i n i t i o n 59.2)

Stokes m u l t i p l i e r s

83

Subdominant s o h ti ons

1 9 ( D e f i n i t i o n 7 . 1 ) , 220(a complete s e t o f ) , 235( Theorem 57.1)

Transcendental s i n g u l a r i t i e s of t h e i n v e r s e of a meromorphic function

173(Remark 1)

Transition points

242(D e f i n i t i o n 59 .l)

bm

1 5 ( d e f i n i t i o n , Theorem 6.1)

bm,k

1 8 ( d e f i n i t i o n , Theorem 7.1), 82(Lemma 21.1)

Zernike s method

53