Solution of some problems of adjoining by reduction to the generalized riemann problem

SOLUTION OF SOME PROBLEMS OF ADJOINING BY REBUCTION TO THE GENERALIZED RIEMANN PROBLEM (RESBENIE NEKOTORYKR ZADACS SOPRXAKRENIIA SVEDKNIRR K OBOBSBCHENNOI ZADACBE RIMANA) PMM

Vo1.27, No.2, 1963, pp.

E.V.

351-355

SKVORTSOV. 8. Kh. FARZAN and A. Ia. CXILAP (Kazan) (Received

July

5,

1962)

In this work, we understand bF the problem of adjoining, the problem of finding in some two-dimensional region D, the solution of the equation div(k grad p) = 0 under the condition that the coefficient k is a piecewise continuous function of the x, y coordinates. On the boundaries y where the coefficient k is discontinuous the following conditions of adjoining must be fulfilled

P+ = P->

k+(g)+=&)_

(1)

Here the subscripts minus and plus denote the limiting values of the function at the boundaries y of discontinuity; II is the normal to y. The solution function p(x, y) has singularities of the logarithmic type in the region D, and if D has a boundary r, the function p(x, y) must satisfy on the r boundary conditions of the first, second, or third kind. This problem has many applications. For example, in underground hydromechanics a function p, which satisfies the conditions given above, determines the pressure field in a piece-wise inhomogeneous oil-containing layer subjected to water pressure and to the linear law of ffltration; in the electromagnetic theory, the function p yields the statistfcal distribution of temperature in a piece-wise nonuniform heat conducting medium. The proposed problem is reduced by the method presented in flf and t21 to a singular integral equation (or a system of equations) for the determination of the solution function p(x, y) on y. The

519

520

Skvortsov,

E.V.

resulting

equation

problem

[41 by the

Hadamard type.

be easily

problems

of

adjoining

on the

takes

on a value p has

2.

value

circle

3.

The region

k,

=

A(q,, k is

(bly

4. tion

=

y<

p)

the the

y
D is

the

region

is

D is

k is point

two

of

solution obtained

of

we shall physics

conduction,

We shall

D_,

k

in which the

6 D,,

func-

a circle

of

<

the

1,

y is

radius;

region

the

+ b2x

the

the

k

point

coefficient

p = 0.

- p
+ cly)

In the

at

Im t < 0)

function

+ blx

yo).

unit

a singularity

- a < x < 6,

k2 = (a2

k = k2 =

where

with the real axis. In the z = x + iy), the coefficient

by (Iz(

In the

singularity

circle,

region

The equa-

< 8.

a f and

x <

P(X,

a, y)

- p f has a

D_(a < x < 6, - p f

+ c~Y)~.

On the

boundary

of

have

problems inclusion usefulness

problems

of

been

in this

solved

before

by methods

article

makes it

and correctness

adjoining

of

our

by comparing

possible

our new approach results

with

earlier.

Let us consider

forms

is

these the

heat

the real axis. in which k

D,,

A(x,,,yo)

this

p has

kl = (al

A(xo, k

demonstrate

of

of

function

x = a.

for

Finally.

y is

a logarithmic

of

coincides Im z > 0,

a rectangle: y is

Their

areas

exterior

D_ given

from ours.

the

concrete

following:

p = 0.

The first

to

the

half-plane

point

and p has

the

interior

different

those

lower

Riemann problem

Among the

work are

Cauchy-

the boundary y is the circumThe region D+ is the interior of

When (~1 = 1,

coefficient

us to

cases.

Riemann

the

plane;

radius.

and the

+ c~)~.

function

entire

= const.

D_

at the

the

0 the

the unit

coefficient

singularity

of

singularity.

k = k,

boundary

generalized

generalized

of

At a certain

circle which by (I z( < 1,

The region of

the

The boundary the upper half-plane

and the

= const.

+ Q2.

(b2y

entire

Chilap

integrals

the obtained

in a number

plane.

8,);

In the

yo). k2

of

A.la.

to of

the

of

where

diameter of this region D,, given is

reduced

in this

D is

A(r,,

and

properties

= const,

k, kg

a circle

at a point const.

in turn the

considered

a logarithmic

of

Farran

two subregions:

The region

ference this

D is

D into

takes tion

of

obtained

The region

1.

divides

then

use

The solution

can then

It

is

R.Kh.

seek

show that

the

and mechanics underground the

first

a function

such

problems as the

of

hydromechanics one of p(x,

y)

the

of

arise

in many

electromagnetism,

and others.

four

in the

adjoining

theories

mentioned

regions

D,

problems. and D_ of

the

Problems

of

the

generalized

$

(2, Y) =

p+

P

Rienann

ydt

(8 (t

_

x)2

521

problem

(2)

G

ya +

+

5p Q)it - ydt

p- (2, Y) = - -+

(3)

%)a+ ya

-co

respectively. axis;

Here p(t)

G is

Green’s

stands

function C=&ln

The quantity

Q is

The solution of the tions (1) identically; following

integral

the

for

for

the unknown function

the

1

upper

(z -

2oY + (Y -

YO12

b -

zoja +

(Y +

YOP

the

source

intensity

of

assumed form satisfies the second one of equation

for

the

s P

4,

For the analytic

solution

0)

of

(t

kl + ks

equation

real

at the

point

A(%,,,

first one of the conditions yields

Q

d”,)P =

this

the these

determination

0;)

on the

half-plane

of

yo).

condithe

p(t):

~34 YO=

@-

we introduce

the

(5)

piece-wise

function

& y

Q, (4 =

P 0) g$

---a)

The limiting

values

of

this

function

and of

its

derivative

have

the

form

Here,

the

integral

Cauchy principal Cauchy-Hadamard the

following

in the

first

formula

or, with the introduction f.&+ -

is

to mean the

it is taken in the (5) is reduced to

of f#ll-

a new function =

the

simplest

The solution

Q?l”

P-

problem is

1

QYO

ni (/cl + k2) (I _

(k, t

nl

This

interpreted

formula equation

Riemann problem:

@‘+ tx) + @‘- tx) =

continuity.

is

while in the second value, sense. With the aid of (6).

of

given

kz) (x -

zo)2 +

ol(z),

(7

yo2

to

the

form

1 ~0)~+ yo*

determining by an integral

a function of

the

by its Cauchy

distype

522

E.V.

Skvortsov,

1

(4 =

Suppose of

of

Cauchy’

Farzan

A.Ia.

ni (kl + b) [(t --zo12

density

9(t)

and

Chilap

4” s

us denote

values

R.Kh.

= q+(t)

integral

that

are

s formula

we have

analytic

and 9-(t)

in D,

@‘1+ (4 = ‘p+ (4 + cp- (a),

4

by q(t).

where 9’(t)

+ 9-(t).

functions

Y021(t

and fl_.

ml- (z) = -

‘p-

are

Then,

the

boundary

on the

basis

(4 + cp- (-1

Since

QY~

cp(4 =

ni

(b

+

1 k2)

(t

-

Q

zoJ2

+

yo2

i

3

t-xo+iyo

1

i

1

2ni (ICI+ k2)

=

t-x0-iy,

-

then

@I+(Z)= since

and,

O’+ = Ol+,

s

P(X) =

Q

on the

PI+ (2, Y) = rt In accordance

the

2n (kl +

formulas

Cauchy’s

Y) =

2n

(2),

1

Direct

verification

In the

second

-

-

(6).

x0J2

that

+

~021

-I-

cd

we have

x012

+

Yo21+ cd

t-z

dt

we obtain

k2)

~0)~+ (Y f YOYI + c

(8) (9)

p- (2, Y) = p.- (2, Y)

the

functions

given

by (8)

and (9)

conditions.

problem

is

[(t

---a,

(k;+ In 1(x-

it

is

convenient i0,

z = rP, The solution

[ix

and (3).

0x 2i (In

formula

shows that

the required

{In

k2)

s

P+ (5, Y) = PI+ (x, Y) + G,

satisfy

2n (/cl +

by formula

Q

x0) dx

~JC(hQf k2) IrnSi

with

PI* h

we find,

x0)2 + yo2 =

of

Q k2) -z - 1 ZO

ml- (2) =

To ’

= - a),-,

(x (2 -

basis

1

k*) z -

O’-

Q

n (kl + k2)

Thus,

-

2r, (k1 +

zo = roe

sought

in the

p+ (rr6) = 5.

c 0

p (4

use

the notation

,

form

2x

1

to

i -

1 - r2 2r cos (0 _ 6) +

r2

da +

G

(10)

of

Problems

the

generalized

Riemann

523

problem

(11)

where rzr$ C=&

The

second

equation

for

condition the

of

-zi

s

P

.

2rro COS (0 2rrO COS (0 -

adjoining

determination

of

2x

1

-

In -yoc

1

leads

=

(

2n (kr + ks)

0

The

integral

Radamard tion

o(z)

The IZI

on the

sense.

For

and

its

limiting

= 1,

left-hand

the

the

following

integral

1is

of

this

as

z

ro2 - 1 2r0 cos ($ - 00) + here

understood

equation

+ i)

r$ in

the

we introduce

(12)

Cauchya func-

derivative

values

are

side

solution

to

1 P

p(o)

Q

(4 dJ

sin” (5 -$)/F

00) + f3O)+

equal

of

o’(z).

approaches

the

circumference

to

Hence 277

1

da

4n

s

9)/Z] = -

sin” I(5 -

p (‘)

(@‘+ (0) + W- (0))

(13)

0

P(N The

equation

(12)

is

=

thus

w+ (0)

Solving

this

reduced

p(o)=Finally,

2n (kr + ks)

problem,

from

P+(r, 0) = -

(0) do (14)

0

Q @‘+ + @‘- =

w-

-

we obtain

(

to

the

Riemann

problem

1 - r,l 2r0 c0s (9 - OO)+

1-

by means

of

(14)

r$

the

-

i)

function

2n (/clQf ka) [In(o--J-)+ln(~--ZO)-lno]+c, (10)

n

and

(11)

we obtain

tklQ+In 1/r%+? k2)

-

2rro cos (O -

eo) + 1 + G + c

p(o)

E-V.

524

P- k‘

8) = -

-$$

R.Kh.

Skuortsou,

tinI’+ c/QQ+

-

ka)

In the problems harmonic functions

Fartan

2PP,

cos

and

(0 -

eo} +

ro= -

ln r) - i&Y&Inr+c

3 and 4 we pass from the functions

a*(~,

p*(s, y) by means of the substitutions

Q

p+1/&T- IL++

2x

p-v7&

Jfk;

The conditions

of adjoining

In problem 3,

the solution,

y) to

u-

(15)

”

where k, is the value of the coefficient

sought

Chila

A.la.

kI at the point

on y for

the functions

which satisfies

A(xo,

yo).

uf take the form

the conditions

(16).

iS

in the form

5 {

IL* (2, y) = f

(4

YU

(qy -

1 %)a + y%-

-1

1

(1 - zz)B+ r%’ 1 d.S

I

u-P, Y) = -

5

s

YU W

-1

Condition

leads

(17)

(T -

now to an integral

s -i l

1

u (4 - z

u (4

--

(T _

1 %)a

(1

-1

1

1 r)Z + ya -

-l

(I-Tz)a++ya

equation

&s 1dT-

Pi

(2 _

is Green’s

function

Let

introduce

us

for

20) (2 ;o) (a _

20-9 G-1)

the semicircle

the piece-wise

analytic

for

function

aZ

(19)

1 d*

U(Z)

ix (z, 20)

Here

G(2,4 = In tz

w

1 i/=0

=

0

(20)

Problems

and its

derivative

of

the

generalized

Riemann

525

problem

a’(z).

equation (2) can be reduced to the With the aid of these functions, generalized Riemann problem of finding a piece-wise analytic function from conditions given on the real axis

ia@+ (x)

@‘+(x) + a’+ (4 +

(0” (xj + iacft- (2) -

= -

-$-CD+ (xj =

-

$- W(x)

cp’- (2) +

afif (2)

for

(+)

+ $f

Isl
for I x I>

1

where f (4

for

=

(x_;o;i+yo.

2iyo

-

1% (a?

+

(20~ +

yo?

!/OS) -

ro1* +

$4

Solving this problem, and making use of the formulas the limiting values of the function (0 on y, we find

of Sokhotskii that. for

IA < 1 u (2) = F (z) -

BXco9 ax -

Ba sin a2

where B

F (1) + F (2 cos a

=

1

co9 ff (z -

F (4

SYO

expx0” +

-

I

yea cosa(z-

PO

C-V

c

a1 (‘r/(x -

(x

~a-1

+

exp

x

2

w0

u. (x -

sin

-

20)

bo’

Y#

cos

a

a”

( V

(+ -

+

II

tar’

-

YO 1

Yo”) + I

+

-

s*sl 20) +

-

z-

Q tx

xola

-4

y”*

$0) +

n

un-’ -

L 6-V

“-_5

b-~0)

60’ + Yo’)’-

Yo

x0’ + yes co9

YO*-

(z-20)*+

n.n! -

+

2 sin a

1/

zo) Ina

20)

x0)*

20)~

u

F PI - F(-

Ba =

’

zO)In 0 V(Z -

- sin a (x +

1)

ZO

310 I

-

aroS I (soa+ y,l,l) n X

(x - 20)boa+ y’o) YO

n

Here. the summation on n (and later on m) is performed from 1 to m. Omitting the details, which are analogous to the preceding ones, we now give the result for the fourth problem

526

E.V.

Skvortsov,

xx

Ll.Kh.

Farzan

I)%

(-

m2 j rJa +

/ p”

a2Pn (~1 +

hr0

+

A.Ia.

Chilap

mrc (~0 + a) 2a

.

$

s’n

4Q (d + CZY)

‘4- (5, Y) =

and

2

I

p4~~+--2)m

~1~0)

I*

x

inn (8 -

sin inn (Y + PI / 31 sin Inn (~0 + P) I281 (d + c2Y)2cochnn (6 - a) _-n

X

C(c + ClY) coth ?+

c +

ClY

33

I 33

4

/ 331

X

e + fy (c + av) Js

l)mm sin ma (% + a) m2/aa + na/ pa 2a 1 (-

xX where

c = a1 + blu,

d =

aa +

bzr,

e =

cbl -dbl,

f = blcl -

hs

BIBLIOGRAPHY 1.

Chilap,

A. Ia.,

K zadache

neodnorodnykh fields gas 2.

in No.

Chilap, plaste layer). Kazan.

3.

Chikin,

4.

1,

polia

problem

nonhomogeneous

of

davlenii

v kusochno-

determining

layers).

IZV.

the

Pressure

UUZOV, Neft’

i

1961.

A. Ia.,

Pole

davlenii

(Pressure

field

Otchetnaia

v kusochno-neodnorodnom in a piece-wise

nauchnaia

sektorial’nom

nonhomogeneous

konferentsia

Kazansk.

sector un-ta

za 1961.

1962. L. A.,

Osobye

nykh

problem

and of

Gakhov,

(On the

piece-wise

integral’ un-ta

ob opredelenii

plastakh

Vol.

113,

sluchai

kraevoi

zadachi

Rimana

uravnenii

(Special

cases

of

Riemann’

singular

integral

equations).

Uch.

No.

F. D. , Kraevye

10,

i singuliarnykh s boundary zap.

value

Kazansk.

1953.

zadachi

(Roundary

Value

Problems).

GIFML,

1958.

Translated

by H.P.T.