On a non-linear boundary value problem in the theory of analytic functions arising in quantum field theory

ON A NON-LINEAR BOUNDARY VALUE PROBLEM IN THE THEORY OF ANALYTIC FUNCTIONS ARISING IN QUANTUM FIELD THEORY* P.

DENCHEV (Moscow)

(Received

In the

use

of

dispersion

representation ary value problem

of

plicit

this

form

sponding

relations

[l] - [2],

problems

to

all

to

that

in particular, arise

in

[3]

In this

considered

we find

the

certain

to

[31.

the

a certain

Mandelstam

non-linear

paper in

In addition

solutions.

obatined

1962)

functions.

similar the

and,

frequently

analytic

type,

for

there

for

those

8 June

bound-

we consider We find

solutions

new class

one

an excorre-

of

solu-

tions. Let us formulate It

is

1)

h(t)

required

our problem. to

find

the

function

h(z)

with

the

following

proper-

ties: is

analytic

with

(-1. l), it has a simple h(m) = A,

* l

the

at all

exception

pole

with

points of

of

the

some real

residue

a < 0;

plane

Points is

h(z)

cut

along

$0

the

> 1,

interval at which

Isol regular at infinity

and

A real;

2)

h(G)

3)

almost

Zh.

uych.

= h(z)

for

all

everywhere

mat.,

3.

z at which on the

No.

4.

cut

h(z)

there

7’71-776,

is

defined;

exist

angular

boundary

values**

1963.

* The function h(z) has the angular boundary value h+(s), (h-(s)) the left (right) at the point s of the contour r if there exists limit

of

paths

which

h(z)

when do not

z

tends

touch

to

s from the

I-. 1043

corresponding

side

on a

along

all

1044

P.

on the

left

h+(s)

which

are

Denchev

summable

and satisfy,

almost

everywhere,

the

relation Im h+ (s) = 1h+ (s) Is. Let us find a method the

an explicit

similar

to

properties

h(<)/($

-

form

that

used

(l)-(3).

Let

Z) and the

contour

for

in

the

[31.

us apply r(e)

solutions

Let the

of

this

problem

be a function

h(z)

Cauchy’s

(see

0)

theorem

to

using

possessing

the

function

figure).

We obtain 1 2ni

s-

I‘(C)

h (f) 5-z

dc = Res (w)+Res

(so)+Res (z), (2)

Res(oo)=--h(co)=-A,

Res (80) =

a Res h (so) = So-2 -

&

’

Res (z) = h (a). On the other 1 2ni

s

h (6) &-zdC=2ni

r(c)

As

hand we have

E

-

l

1

s h(o+

-1

is)

do

a+ie-2

1

s h (a -

1

-2ni

is)

1

a-ie-zddO+2ni,

c

-1

(3)

hod& 5-e

Yl+Yr

0 we obtain*

IImm

1

s

h (6)

6--a

1 -d52ni

r(t) Here we are

using

s l

h+b)----(a)

do=

(Z),

In h+ (a) Q--e dc*

-1

-1

property

s

l

1 fl

c-2

from which

it

follows

that

h-(s)

exists

and h-(s) = h+ (9). From (2)

we obtain Im h+ (a) 6--z dd.

l

We assume that

the

can be made in

(3).

function

h(z)

is

such

that

passages

(4)

to the

limit

The

theory

of

analytic

1045

functions

Using (4) we find

Imh(4 =

w I” o _ z do Im 2 = SI, h+ 1

(-*++

la

Since a < 0, P(z) > 0; hence if has no non-real zeros. Let us introduce

Im z # 6, Im hfr)

Im 2.

(5)

f 0 also,

SO that

h(zf

the function

H(2j = -

&.

in the plane The function H(z), analytic since h(s) has no non-real non-real poles, the real axis. At the point se the function infinity H(Z) is regular and if A # 0, H(m) has a pole at infinity. It follows

P (2)

-1

from condition

09 with cut ( - 1, l), has no zeros, but can have Poles on H(z) has a simple zero. At = - l/A; if A = 0 then H(z)

(2) that H(i)

=

--

H (2).

(7)

On the cut there exist angular boundary values H+(s) at all points at From the uniqueness theorem for which h+(s) exists and is non-zero. analytic functions (see [4]) it follows that the set where h+(s) = 0 has zero measure, so that H+(S) exists almost everywhere. From (1) we find that almost everywhere Im H+ (s) = 1. From (5)

(8)

and (6) we obtain

(9) The last

relation

shows that

(f0) It follows from (7) and (10) that H(r) is an R-function* has an integral representation (see [41, p. 117)

l

R-functions half-planes

and therefore

are functions which are regular in the upper and lower and which possess properties (7) and (10).

1046

P. Dencheu

H (3 = PZ + v +

where U. v are real constants, tion of bounded variation. From the Stieltjes-Perron

s

OJi+az 6-_z

CI> 0 and a(o)

conversion

(4%)

da 6%

is a non-decreasing

formula

(see

func-

[51) we have

where

9 (a) =

!(I + +)

da (T)

n

for any real

c and o,

Obviously y(o) is not decreasing. exists almost everywhere and

It follows

from (12) that ~(0)

9,‘(g) = 4 Im H+ (a).

(13)

Outside the cut Im H*(a) = 0 (since H(u) is regular the cut Im H*(o) is given by formula (6). Thus

q (a) =

almost everywhere

f

0

outside

We can decompose the non-decreasing

(-1,

and (7)

on (-1,

l),

v(o)

into

holds)).

(14)

1). function

On

three

terms:

where Q(o) is a completely continuous fUnCtiOn, r(o) is a jUmP function and 6(u) a non-decreasing singular function. i.e. a non-constant continuous function with finite variation, whose derivative is almost everywhere equal to zero (see Es]).

The

theory

of

analytic

1047

functions

Here

(16)

r(a)=Zrke(O--k), k

where E(a)

‘k

>o,

Ck

are

Obviously

1 for

Go;

real.

0(a)

Substituting

for

0

=

= const. in

(11)

outside

( - 1,

1).

we obtain

H (4 = P (4 + y t- I(z) + R(z) + T (~1.

(17)

where

(18) -1

(y(z)

is

the

angle

at which

the

segment

( - 1,

1)

is

seen

from the

point

z) I

T(z)=‘sss.

-1 Since finally

H(Z),

Z(z). R(z), T(Z)

are

bounded

at

infinity,

p = 0 and

we have 1 i-s H (4 = v + 2 ln ps I

Let us show that all the points ( - 1, 1). For, on the real axis for

I

+ ir (z)ln+R

contained in the ISI > 1 we have

Ck,are

@[I

(4 + T (4. interval

1048

P. Dencheu

H (s) =

At infinity

H(S)

is

4 In

Y +

continuous,

(22)

since 1

lim H(s)=v-_$_ s-N*00 Further,

it

is

clear

from

(22)

‘k that

H(-.I--O)~foo.

H(l+O)=-moo,

If + =‘, 1.

(-

there H(ck

+

1):

is

a point

0)

=

one to

h(z)

must have

only

at the

lie

in the

at

point

se.

and (21)

condition

the

of

This

interval

has therefore

1,

the

1) then

H(ck - 0)

two zeros

right

of

(-1,

l),

while

shows

that

outside

contradiction

(-

at least

Ck and one to

two poles

( - 1,

all

=

outside

Ck. But then h(z) the

has a pole

points

Ck

1).

we obtain

(v+ i

h (z) = From the

left

least

interval

From (6)

Ck outside

- a~ and H(S) the

\ w. -1

h(m)

In 1 i<

1 i- k

y (z) + R (2) + T (z))-’

(23)

.

= A we have 1

++++

\

k to the conditions so and so at this point

of the

According point

$$o.

(24)

-1

the problem denominator

at the

h(z) has a pole in (23) becomes

equal

to

zero:

v+,ln

Let

l

1

us calculate

The term

T(Z)

(25)

the

residue

was not

of

taken

h(z)

into

at

so:

account

in the

analysis

in

[31.

The

By hypothesis,

this

theory

of

residue

is

analytic

equal

1049

functions

to

and we obtain

a,

the

relation

(26) We have thus if

is

h(z)

(23),

obtained

the

a solution

where y(z),

of

R(Z),

a non-decreasing satisfying

following our

the

problem are

T(z)

singular

result: it

found

function,

relations

v.

(24)-(26)

can be represented from rk,

(16),

(19),

Ck are

real

by formula (20):

8(o)

is

constants

and (27)

Now let the

us show that

given

conditions

Let us consider from the

all

are the

formula

that

the

functions

solutions

function

defined

by formula

analytic

by formula

everywhere

of

(21).

(3).

in the

( - 1. 1) and can have a singularity only at points in (23) is equal to zero. This denominator is the H(z)

(23)

with

our problem. defined

h(z)

is

h(z)

of

It

plane

is

clear

with

cut

where the denominator same as the function

We have

Im H (z) = -1

It

is

only

for

values

clear

real

z,

1 Z[

> 1.

increase.

Hence there

fore

has only

increases

is

and from

(24)

it

(1)

Condition

are (2)

The existence boundary

properties

(-

one,

It 1,

is

that

h(m)

Im H(z)

seen, I(1

one,

zero (25)

pole that = A.

zero

only

outside

is -

zeros.

f 0

Im H(z)=0 for

these

(22).

continuous

1) as

and only at this is clear

to

by formula

The relation

one pole.

follows

can be equal

as we have

outside

correspondingly,

has no non-real

given

(22).

at the point so. The residue see from (26). From (23) it condition

is

( Ckj < 1 and,

it

H(z)

Thus H(Z)

function

Moreover, h(z)

Im z # 0 then,

them H(z)

the

since

if

z) > 0 and thus

Z, and for

us study l),

that

y(z)Im(

of

Let ( - 1,

thus

since

also,

the

continuous

interval

at infinity.

s)/(l

+ s) 1, T(s). R(s) outside ( - 1, 1); there-

shows

that

this

pole

lies

is

equal to a, as we can h(z) is regular at infinity,

Thus all

the

requirements

of

satisfied. can be verified

directly.

of

everywhere

h+(s) of

the

almost

Cauchy-Stieltjes

on the type

cut

integral

follows (see

from the [41),

and

P.

1050

h+ (s)

This

formula

gives

Thus the ties

(l)Finally, all

with It

the the

result

solutions

Thus for

In (4) find

let

that

If

the

given

is

satisfied

to

(26)

possesses

can be given

that

a solution

-

(9; -

$

these

to

(23)

the proper-

our problem.

as follows:

have

by formula

(23)

it

the

real

the

inequalities (29)

are not

satisfied,

has no solution

point

and the

a < 0 and

1) < u < 0.

inequalities

In particular,

for

s from above.

Sokhotskii

formula

the problem

a = 0 and A # 0.

Then,

since

h+(s)

(see

[41)

holds,

everywhere,

where

+f i

z,

u(s),

U(S)

the

real

and imagin-

da,

are

h+(s).

We can write

condition

(I)

in the

form

V (8) = UP(S)+ u2(s). Thus,

if

functions

we

relation

almost of

of

problem

u(s)=A-_&ary parts

(28)

above.

everywhere

the

by formula

a solution

relation

problem

z tend

almost

exists

of

listed

must be satisfied. has no solution.

+ R (4 + T (4)-l.

can be formulated

from the

the

defined

therefore

conditions

follows

v +

h(z)

and is the

fIni&

(

-

(1).

function

(3)

S

Dencheu

h(z)

is

a solution

and V(S)

U(S)

satisfy

of

the

boundary

the

system

the

converse.

of

(31) value

problem.

equations

(30).

then

the

(31)

almost

everywhere. It of

is

system

not

difficult

(30).

(31)

to then

prove the

function

If

u(s).

v(s)

is

a solution

theory

The

of

1051

+-&$

&

h (z) = A -

functions

analytic

-1 is

a solution

and the thus

of

system

found

the

of

an explicit

In particular, of

boundary

value

equations form

as

for

have

we

The boundary

problem. (31)

(30), all

are

the

seen,

it

therefore

solutions

follows

value

problem

equivalent. of

from

this

(29)

We have

system. that

the

system

equations 1

s

v (4

u (s) = A + f

&

’

0-s

-1

vz(s)

v (s) = 24”(s) + has no solution Let tion

A # 0.

be a solution

h(z)

h+(s)

if

= U(S)

of

24(s) =

-

the

It

+ iv(s).

given

follows

problem. from

L (4 L2

(s)

+

1

(28)

v(4 =

9

Let

us examine

the

func-

that 1

II2 (*s) +

1

’

where i-s In 1$-S + R (-7) + T (8).

L (s) = v + + First

it

follows

Ih+(s)lei6(s)

from

from

(1)

(1)

that

h+(s)

is

bounded.

(32) For

if

h+(s)

=

we have

1h+ (s) 1= sin 6 (s). Let the

Gl denote

derivative

the

set

of

points

of

the

singular

8)

be an interval,

ck,

and C2 the

function

8(a)

both

ends

does

set

of

not

exist

points or

is

at which non-

belong

to

Cl

U Gz

zero. Let

n = (a,

and such

that

Then clearly

the

intersection

A f’ (Cl

of

which

U GzJ is

empty.

1052

p. Denchev

and since L(s) is continuous in (a, p) there exists at least one point in this interval at which L(s) is equal to zero, At this point u(s) is equal to zero, and U(S) is equal to one, as we can see from (32). On the other hand it is clear that at the points c( and fi the functions u(s) and u(s) are equal to zero. Since V(S) is continuous in (a. P) there will be at least two points at which it takes the value !& At these points u(s) takes the values i %. as we can see from (31). It follows from our argument that if the set GI U Gz is such that there exists some point in each neighbourhood of which there is an interval like (a, p) (i.e. whose end-points belong to Gl U G2 and inside which there are no points of G1 u G2) then the functions U(S) and V(S) also h+(s)) will be discontinuous at this point. In (and, therefore, particular, h+(s) will be discontinuous if GI consists of an infinite set of isolated points and G2 is a closed set.

Translated

by R. Feinstein

REFERENCES 1.

Crew, G.F. and Mandelstam,

2.

Efremov, Nucl.

A.V., Phys.,

S.,

Meshcheryakov, 22,

No.

L., Dalitz, 1956.

2,

Phys.

V.A.,

202-206,

Rev.,

119,

Shirkov,

No.

1, 467-477,

1960.

D.V. and Tzu, H.Y.,

1961.

R.H. and Dyson. F.J..

Rev.,

101, No. 1,

3.

Castillejo, 453-458,

4.

Privalov. I. I., ary Properties

5.

problems momentov (The Classical Akhiezer. N. I., Klassicheskaya Problem of Moments). Fizmatgiz, Moscow. 1961.

6.

veshchestvennoi peremennoi (The Natanson. I. P. , Teoriya funktsii Theory of Functions of a Real Variable). Fizmatgiz, Moscow, 1957.

Phys.

Granichnye svoistva analiticheskikh of Analytic Functions). Fizmatgiz,

funktsii (BoundMoscow, 1960.