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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+=
"$-A
0)
.
N=o m
-N-1 f l ( x ) 5 I: (-N)cNx N-
where Proof:
fl(x)
(x+m
denotes t n e d e r i v a t i v e of
f(x)
&
fit)
,
with respect t o
x
.
Let us put N
P,(X) =
z
ChX
-h
h=o
and = f ( x ) -pN(x) We must show t h a t
-N-2)
%(x) = O ( x
s u b s e c t o r of t h e open s e c t o r sector
9
.
gl
'
as
x
tends t o i n f i n i t y i n a closed
To do t h i s , l e t us c o n s i d e r a closed
defined by
where a [$ 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($) 1-l
This completes the proof of Lemma 10.1. W e s h a l l now complete t h e proof of Lamma 9.1 by using Lemma 10.1.
do t h i s , consider a s e c t o r 8 i n the
( al,
...,a m )-space
? - p l a n e and a domain
To
;Qr i n the
which a r e defined r e s p e c t i v e l y by 8: l a r g ho+ (mt2)arg
53 ,
51
151 2n
and pr:
lall t
...+(%I
m-1
(i)
cpk
i s a polynomial i n
(al
i s a polynomial i n
..., N - m t l )
( h = 0 , 1 , **.,m-2)
,...,am,Xo, ...,$-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 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 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_arg 7 2 , a r g A Therefore, we can change -1 2 T=A t
T
by (O ~ 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 = 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, 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 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= 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)
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 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 - 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 (OlSo
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
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
