On solvability of a boundary value problem for singular integral equations

Southeast Asian Bulletin of Mathematics 2002 26: 235–244 Southeast AsianBulletin of Mathematics : Springer-Verlag 2002 On Solvability of a Boundary Value Problem for Singular Integral Eq

Southeast Asian Bulletin of Mathematics (2002) 26: 235–244 Southeast Asian

Bulletin of Mathematics

: Springer-Verlag 2002

On Solvability of a Boundary Value Problem for

Singular Integral Equations

Nguyen Van Mau and Nguyen Tan Hoa

Hanoi University of Science, Vietnam, 334 Nguyen Trai, Dong Da, Hanoi, Vietnam

AMS Subject Classification: 47G05, 45GO5, 45E05

Abstract.This paper deals with the solvability of boundary value problems for singular in-tegral equations of the form (i)–(ii)

By an algebraic method we reduce the problem (i)–(ii) to a system of linear algebraic equations which gives all solutions in a closed form

Keywords: initial and co-initial operators, singular integral equations, boundary value problem

Introduction

The theory of general boundary value problem induced by right invertible oper-ators were investigated by Przeworska-Rolewicz and has been developed by many other mathematicians (c.f [2], [4]) In this paper, we give an application of this theory to solve the following boundary value problem:

½ðKnþ Kn1KcÞjðtÞ ¼ f ðtÞ; ðiÞ

ðFjKjjÞðtÞ ¼ jjðtÞ; j¼ 0; ; n  1; jjðtÞ A Ker K; if k> 0;

ðG0fÞðtÞ ¼ 0; ðGjRj1 .R0Þ f ðtÞ ¼ 0; j¼ 1; ; n  1 if k< 0; ðiiÞ where Kc is an operator of multiplication by the function cðtÞ; Fj; Gj ð j ¼ 0; ;

n 1Þ are initial and co-initial operators, respectively, and

ðKjÞðtÞ ¼ aðtÞjðtÞ þbðtÞ

pi ð

G

jðtÞ

t tdt; k¼ Ind K:

1 Preliminaries

We recall some notations and results which are used in the sequel (see [2], [4])

Let G be a simple regular closed arc in complex plane and let X ¼ HmðGÞ ð0 < m < 1Þ Denote by Dþ the domain bounded by G (assume that 0 A Dþ) and

by D-its complement including the point at infinity The set of all linear operators with domains and ranges contained in X will be denoted by LðX Þ Write L0ðX Þ :¼

fA A LðX Þ : dom A ¼ X g

Let RðX Þ be the set of all right invertible operators belonging to LðX Þ For

D A RðX Þ, we denote by RDthe set of all its right inverses

An operator F A LðX Þ is said to be an initial operator for an operator D A RðX Þ corresponding to a right inverse R of D if

F2 ¼ F ; Fðdom DÞ ¼ Ker D; FR¼ 0:

Denote by FD the set of all initial operators for D A RðX Þ

The following fact is well-known:

F A LðX Þ is an initial operator for D A RðX Þ corresponding to R A RD if and only if F ¼ I  RD on dom D

Let LðX Þ be the set of all left invertible operators belonging to L0ðX Þ For

V A LðX Þ, we denote by LV the set of all its left inverser

If V A LðX Þ and L A LV then the operator

G :¼ I  VL

is called the co-initial operators for V corresponding to L A LV

Denote by GV the set of all co-initial operators for V A LðX Þ

Lemma 1 (see [2]) Let A; B A LðX Þ, Im A H dom B, Im B H dom A Then equa-tion ðI  ABÞx ¼ y hassolutionsif and only if ðI  BAÞu ¼ By doesand there is one-to-one correspondence between the two sets of solutions, given by

u¼ Bx $ x ¼ y þ Au:

2 The Results

Let

ðKjÞðtÞ :¼ aðtÞjðtÞ þbðtÞ

pi ð

G

jðtÞ

t tdt;

where aðtÞ; bðtÞ A X , a2ðtÞ  b2ðtÞ ¼ 1, 0 0 k ¼ Ind K

Denote

ðR0jÞðtÞ :¼ aðtÞjðtÞ bðtÞZðtÞ

pi ð

G

jðtÞ ZðtÞðt  tÞdt;

where

ZðtÞ ¼ eGðtÞtk=2; GðtÞ ¼2pi1

ð

G

ln tk aðtÞbðtÞ aðtÞþbðtÞ

t t dt:

It is known that (see [2], [4])

i) If k> 0 then K is right invertible and

RK¼ fR ¼ R0þ ðI  R0KÞT : T A L0ðX Þg:

Let F0; ; Fn1be given initial operators for K corresponding to R0; ; Rn1A

RK, where

Rj¼ R0þ ðI  R0KÞTj ð j ¼ 1; ; n  1Þ; T1; ; Tn1 AL0ðX Þ: ð1Þ ii) If k< 0 then K is left inverbible and

LK¼ fR ¼ R0þ TðI  KR0Þ : T A LðX Þ; dom T ¼ ðI  KR0ÞX g:

In this case, let G0; ; Gn1be given co-initial operators for K corresponding to

R0; ; Rn1A LK, where

Rj¼ R0þ TjðI  KR0Þ ð j ¼ 1; ; n  1Þ; T1; ; Tn1ALðX Þ: ð2Þ Lemma 2.If k> 0, then

ðF0jÞðtÞ ¼X

k1

k¼0

ukðjÞckðtÞ on X ;

where ckðtÞ ¼ bðtÞZðtÞtk ðk ¼ 0; ; k  1Þ and ukðjÞ ðk ¼ 0; ; k  1Þ are linear functionalswhich are defined by

ukðjÞ ¼ 1

2pi ð

G

tk1k

eGðtÞ jðtÞ  1

pi ð

G

jðt1Þ

t1 tdt1

where GðtÞ isa boundary value of the function GðzÞ in D

Proof.We have

ðF0jÞðtÞ ¼ ½ðI  R0KÞjðtÞ

¼ jðtÞ  a2ðtÞjðtÞ aðtÞbðtÞ

pi ð

G

jðtÞ

t tdt

þbðtÞZðtÞ pi ð

G

1 ZðtÞðt  tÞ aðtÞjðtÞ þ

bðtÞ pi ð

G

jðt1Þ

t1 tdt1

dt

¼ jðtÞ  a2ðtÞjðtÞ aðtÞbðtÞ

pi ð

G

jðtÞ

t tdt

þ bðtÞZðtÞ 1

pi ð

G

jþðtÞ dt

XþðtÞðt  tÞ

1 pi ð

G

jðtÞ dt

XðtÞðt  tÞ

;

On Solvability of a Boundary Value Problem for Singular Integral Equations 237

where jþðtÞ ¼1

2½ðI þ SÞjðtÞ, jðtÞ ¼1

2½ðI þ SÞjðtÞ, XþðtÞ ¼ eG þ ðtÞ, XðtÞ ¼

tkeGðtÞ

(GþðtÞ; GðtÞ are boundary values of the function GðzÞ in Dþ; D, re-spectively)

On the other hand

1

pi

ð

G

jþðtÞ dt

XþðtÞðt  tÞ

1 pi ð

G

jðtÞ dt

XðtÞðt  tÞ

¼ j

þðtÞ

XþðtÞ

1 pi ð

G

jðtÞtkdt

eG  ðtÞðt  tÞ

¼ j

þðtÞ

XþðtÞ

tk

pi ð

G

jðtÞ dt

eGðtÞðt  tÞ

Xk1

k¼0

tk

pi ð

G

tk1kjðtÞ

eGðtÞ dt

¼ j

þðtÞ

XþðtÞþ

jðtÞ

XðtÞ

Xk1

k¼0

tk

pi ð

G

tk1kjðtÞ

eG  ðtÞ dt

¼bðtÞ

ZðtÞjðtÞ þ

aðtÞ ZðtÞpi ð

G

jðtÞ

t tdt

Xk1

k¼0

tk

pi ð

G

tk1kjðtÞ

eG  ðtÞ dt:

Hence

ðF0jÞðtÞ ¼X

k1

k¼0

bðtÞZðtÞtk

2pi ð

G

tk1k

eGðtÞ jðtÞ  1

pi ð

G

jðt1Þ

t1 tdt1

dt:

By similar arguments, we obtain the following result:

Lemma 3.If k< 0, then

ðG0jÞðtÞ ¼ X

jkj1

k¼0

vkðjÞckðtÞ on X;

where ckðtÞ ¼ bðtÞtk ðk ¼ 0; ; jkj  1Þ and vkðjÞ ðk ¼ 0; jkj  1Þ are linear functionalsdefined by

vkðjÞ ¼ 1

2pi ð

G

tjkj1keGðtÞ jðtÞ

ZðtÞ

1 pi ð

G

jðt1Þ dt1

Zðt1Þðt1 tÞ

where GðtÞ isa boundary value of the function GðzÞ in D

In the sequel, for every function cðtÞ A X , we write

ðKjÞðtÞ ¼ cðtÞjðtÞ:

Consider singular integral equation of the form

½ðKnþ Kn1KcÞjðtÞ ¼ f ðtÞ; ð5Þ with mixed boundary conditions

iÞ ðFjKjjÞðtÞ ¼ jjðtÞ; jjðtÞ A Ker K; j¼ 0; ; n  1 if k> 0;

iiÞ ðG0fÞðtÞ ¼ 0; ðGjR0 .Rj1fÞðtÞ ¼ 0; j¼ 1; ; n  1 if k< 0; ð6Þ where fðtÞ; cðtÞ A X ; Fj; Gjð j ¼ 0; ; n  1Þ are defined by (1) and (2), respec-tively, 1< n A N

Theorem 1.Suppose that1þ cðtÞaðtÞ G cðtÞbðtÞ 0 0 for all t A G Then every solu-tion of the problem(5)–(6) can be found in a closed form

Proof

Let k> 0, we have K A RðX Þ

Hence, the equation (5) is equivalent to the equation

jðtÞ ¼ ðR0 .Rn1Kn1KcjÞðtÞ þ ðR0 .Rn1fÞðtÞ þ ðR0 .Rn2zn1ÞðtÞ

þ þ ðR0z1ÞðtÞ þ z0ðtÞ;

where z0ðtÞ; ; zn1ðtÞ A Ker K are arbitrary

Thus, the problem (5)–(6) is equivalent to the equation

jðtÞ ¼ ðR0 .Rn1Kn1KcjÞðtÞ þ ðR0 .Rn1fÞðtÞ þ ðR0 .Rn2jn1ÞðtÞ

þ þ ðR0j1ÞðtÞ þ j0ðtÞ;

i.e

½ðI þ R0 .Rn1Kn1KcÞjðtÞ ¼ f1ðtÞ; ð7Þ where

f1ðtÞ ¼ ðR0 .Rn1fÞðtÞ þ ðR0 .Rn2jn1ÞðtÞ þ þ ðR0j1ÞðtÞ þ j0ðtÞ:

By the Taylor-Gontcharov formula for right invertible operators (see [4]), (7) is iquivalent to the equation

2

4

0

@Iþ R0 I F1X

n1

k¼2

R1 .Rk1FkKk1

!

Kc

1 Aj

3 5ðtÞ ¼ f1ðtÞ: ð8Þ

By lemma 1, in order to solve the equation (8) it is enough to solve the equation

On Solvability of a Boundary Value Problem for Singular Integral Equations 239

Iþ KcR0 F1KcR0X

n1

k¼2

R1 .Rk1FkKk1KcR0

! c

ðtÞ ¼ gðtÞ;

where

gðtÞ ¼ I F1X

n1

k¼2

R1 .Rk1FkKk1

!

Kcf1

ðtÞ:

Rewrite this equation in the form

Iþ KcR0 F1KcR0X

n1

k¼2

R1 .Rk1FkKk1KcR0

!

KZf

ðtÞ ¼ gðtÞ; ð9Þ

where fðtÞ ¼ cðtÞ=ZðtÞ

From lemma 2, we have

ðFkKk1KcR0KZfÞðtÞ ¼X

k1

j¼0

ujkðfÞcjðtÞ; k¼ 1; ; n  1;

where ujkðfÞ ¼ uj½ðI  TkKÞKk1KcR0KZf; ujðjÞ; cjðtÞ ð j ¼ 0; ; k  1Þ are defined by (3)

Hence, (9) is of the form

ðMfÞðtÞ X

n1

k¼1

Xk1

j¼0

ujkðfÞcjkðtÞ ¼ gðtÞ; ð10Þ

where cj1ðtÞ :¼ cjðtÞ, cjkðtÞ ¼ ðR1 .Rk1cjÞðtÞ ðk ¼ 2; ; n  1Þ, j ¼ 0; ;

k 1 and

ðMfÞðtÞ :¼ ½1 þ cðtÞaðtÞZðtÞfðtÞ cðtÞbðtÞZðtÞ

pi ð

G

fðtÞ

t tdt: ð11Þ Write this equation in the form

ðMfÞðtÞ X

q

k¼1

~

ukðfÞ ~ckðtÞ ¼ gðtÞ; ð12Þ

where q¼ kðn  1Þ, f~u1ðfÞ; ; ~uqðfÞg is a permutation of fujkðfÞ; j ¼ 0; k  1;

k¼ 1; ; n  1g and f ~c1ðtÞ; ; ~cqðtÞg is obtained by this permutation from the set of functionsfc ðtÞ; j ¼ 0; ; k  1; k ¼ 1; ; n  1g

ðNfÞðtÞ :¼½1 þ aðtÞcðtÞfðtÞ þ

cðtÞbðtÞZ 1 ðtÞ pi

Ð

G fðtÞ

Z 1 ðtÞðttÞdt

1 þ aðtÞcðtÞ2 c2ðtÞb2ðtÞZðtÞ ; where

Z1ðtÞ ¼ ZðtÞeG1 ðtÞtk1 =21 þ aðtÞcðtÞ2 b2ðtÞc2ðtÞ1=2

;

G1ðtÞ ¼ 1

2pi ð

G

lntk 1GðtÞ

t t dt;

GðtÞ ¼1þ aðtÞcðtÞ þ bðtÞcðtÞ

1þ aðtÞcðtÞ  bðtÞcðtÞ; k1¼ Ind GðtÞ:

If k1¼ 0, then M is invertible and M1¼ N Hence, the equation (12) is equivalent to the equation

fðtÞ X

q

k¼1

~

ukðfÞðN ~ckÞðtÞ ¼ ðNgÞðtÞ: ð13Þ

Without loss of generality, we can assume that fðN ~ckÞðtÞgk¼1; q is a linearly independent system Then every solution of (13) can be found in a closed form by means of the system of linear algebraic equations

~

ujðfÞ X

q

k¼1

ajku~kðfÞ ¼ ~ukðNgÞ; j¼ 0; ; q;

where ajk¼ ~ujðN ~ckÞ; k; j ¼ 1; ; q

If k1> 0, then M is right invertible and N is a right inverse of M Hence, the equation (12) is equivalent to the equation

fðtÞ X

q

k¼1

~

ukðfÞðN ~ckÞðtÞ ¼ ðNgÞðtÞ þ yðtÞ;

where yðtÞ A Ker M is arbitrary

We now can solve this equation by the same method as for the equation (13), i.e every its solution can be found in a closed form

If k1< 0, then M is left invertible and N is a left inverse of M Hence, the equation (12) is equivalent to the system

fðtÞ X

q

k¼1

~

ukðfÞðN ~ckÞðtÞ ¼ ðNgÞðtÞ;

ð

G

gðtÞ þP

q k¼1

~

ukðfÞ ~ckðtÞ

Z1ðtÞ t

n1dt¼ 0; n¼ 1; ; jk1j;

8

>

>

<

>

>

:

On Solvability of a Boundary Value Problem for Singular Integral Equations 241

fðtÞ X

q

k¼1

~

ukðfÞðN ~ckÞðtÞ ¼ ðNgÞðtÞ;

Xq

k¼1

bnku~kðfÞ ¼ fn; n¼ 1; ; jk1j;

8

>

>

>

>

ð14Þ

where

fn¼  ð

G

gðtÞtn1dt

Z1ðtÞ ; bnk¼

ð

G

~

ckðtÞtn1

Z1ðtÞ dt:

Without loss of generality, we can assume that fðN ~ckÞðtÞgk¼1; q is a linearly independent system Every solution of (14) can be found in a closed form by means of the system of linear algebraic equations

~

ujðfÞ X

q

k¼1

ajku~kðfÞ ¼ ~ujðNgÞ; j¼ 1; ; q;

X

q

k¼1

bnku~kðfÞ ¼ fn; n¼ 1; ; jk1j;

8

>

>

>

>

where akj¼ ~ukðN ~cjÞ, k; j ¼ 1; ; q

Thus, every solution of the equation (12) can be found in a closed form Due to the result of Lemma 1, every solution of the problem (5)–(6) is defined

by the formula

jðtÞ ¼ ðR0KZfÞðtÞ þ f1ðtÞ;

where fðtÞ is a solution of the equation (12), i.e every solution of the problem (5)– (6) can be found in a closed form

Let k< 0, we have K A LðX Þ

The Taylor-Gontcharov formula for left invertible operators (see [4]) and (6) together imply

½ðI  KnRn1 .R0Þ f ðtÞ ¼ ðG0fÞðtÞ þ X

n1

k¼1

KkGkRk1 .R0

! f

ðtÞ ¼ 0; i.e

fðtÞ ¼ ½ðKnRn1 .R0Þ f ðtÞ:

Hence, the problem (5)–(6) is equivalent to the equation

½ðKnþ Kn1KÞjðtÞ ¼ ðKnRn1 .R fÞðtÞ;

½ðK þ KcÞjðtÞ ¼ ðKRn1 .R0fÞðtÞ: ð15Þ

If jðtÞ is a solution of (15) then

ðG0KcjÞðtÞ ¼ ðG0KRn1 .R0fÞðtÞ  ðG0KjÞðtÞ ¼ 0;

Thus, (15) is equivalent to the system

½ðI þ R0KcÞjðtÞ ¼ f2ðtÞ;

ðG0KcjÞðtÞ ¼ 0;

ð16Þ

where f2ðtÞ ¼ ðRn1 .R0fÞðtÞ

From Lemma 3,ðG0KcjÞðtÞ ¼ 0 if and only if

vkðKcjÞ ¼ 0; k¼ 0; ; jkj  1;

where vkðjÞ ðk ¼ 0; ; jkj  1Þ are defined by (4)

Hence, the system (16) is equivalent to the system

½ðI þ R0KcÞjðtÞ ¼ f2ðtÞ;

vkðKcjÞ ¼ 0; k¼ 0; ; jkj  1:

Consider the system of equations

½ðI þ KcR0ÞcðtÞ ¼ ðKcf2ÞðtÞ;

vkðcÞ ¼ 0; k¼ 0; ; jkj  1:

ð17Þ

It is easy to check that the system (16) has solutions if and only if the system (17) does Moreover, if jðtÞ is a solution of (16) then cðtÞ ¼ cðtÞjðtÞ is a solution of (17) Conversely, if cðtÞ is a solution of (17) then

jðtÞ ¼bðtÞcðtÞ þ

bðtÞZðtÞ pi

Ð

G cðtÞ ZðtÞðttÞdtþ f2ðtÞ

1þ cðtÞaðtÞ þ cðtÞbðtÞ ð18Þ

is a solution of (16)

Hence, in order to solve the system (16), it is enough to solve the system (17) Rewrite (17) in the form

ðMfÞðtÞ ¼ ðKcf2ÞðtÞ;

~

vvkðfÞ ¼ 0; k¼ 0; ; jkj  1:

ð19Þ where fðtÞ ¼ cðtÞ=ZðtÞ, ~vv ðfÞ ¼ v ðK fÞ and M is defined by (11)

On Solvability of a Boundary Value Problem for Singular Integral Equations 243

By the same method as for the equation (12), every solution of this system can

be found in a closed form So every solution of the problem (5)–(6) is defined by the formula

jðtÞ ¼bðtÞZðtÞfðtÞ þ

bðtÞZðtÞ pi

Ð

G

fðtÞ ðttÞdtþ f2ðtÞ

1þ cðtÞaðtÞ þ cðtÞbðtÞ ; where fðtÞ is a solution of the system (19), i.e every solution of the problem (5)– (6) can be found in a closed form

References

1 Gakhov, F.D.: Boundary value problems, Oxford, 1966 (3rd Russian complemented and corrected edition, Moscow, 1977)

2 Mau, Ng.V.: Boundary value problems and controllability of linear system with right invertible operators, Warszawa, 1992

3 Mau, Ng.V.: Generalized algebraic elementsand linear singular integral equationswith transformed arguments, WPW, Warszawa, 1989

4 Przeworska-Rolewicz, D.: Algebraic Analysis, PWN and Reidel, Warszawa-Dordrecht, 1988

5 Przeworska-Rolewicz, D.: Equationswith transformed argument, An algebraic approach, Amsterdam-Warszawa, 1973

By lemma 1, in order to solve the equation (8) it is enough to solve the equation

On Solvability of a Boundary Value Problem for Singular Integral Equations 239... problems and controllability of linear system with right invertible operators, Warszawa, 1992

3 Mau, Ng.V.: Generalized algebraic elementsand linear singular integral equationswith transformed... Integral Equations 243

By the same method as for the equation (12), every solution of this