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Complex Variables and Elliptic
Equations
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On a class of singular integral equations with the linear
fractional Carleman shift and the degenerate kernel
Le Huy Chuan a; Nguyen Van Mau b; Nguyen Minh Tuan b
a
Department of Applied Physics, Graduate School of Engineering, Osaka
University, Japan
b
Faculty of Mathematics Mechanics and Informatics, Department of Analysis,
University of Hanoi, Hanoi, Vietnam

Online Publication Date: 01 February 2008
To cite this Article: Chuan, Le Huy, Van Mau, Nguyen and Tuan, Nguyen Minh
(2008) 'On a class of singular integral equations with the linear fractional Carleman shift and the degenerate kernel',
Complex Variables and Elliptic Equations, 53:2, 117 - 137
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Complex Variables and Elliptic Equations
Vol. 53, No. 2, February 2008, 117–137

On a class of singular integral equations with the
linear fractional Carleman shift and the
degenerate kernel
LE HUY CHUANy, NGUYEN VAN MAUz and NGUYEN MINH TUAN*z
yDepartment of Applied Physics, Graduate School of Engineering, Osaka University, Japan
zFaculty of Mathematics Mechanics and Informatics, Department of Analysis,
University of Hanoi, 334, Nguyen Trai Str., Hanoi, Vietnam
Communicated by R. P. Gilbert
(Received 8 March 2007; in final form 9 June 2007)
This article deals with the solvability, the explicit solutions of a class of singular integral
equations with a linear-fractional Carleman shift and the degenerate kernel on the unit circle
by means of the Riemann boundary value problem and of a system of linear algebraic
equations. All cases about index of the coefficients in the equations are considered in detail.
Keywords: Integral operators; Singular Integral equations; Riemann boundary value problems
AMS Subject Classifications: 47G05; 45G05; 45E05

1. Introduction

Singular integral equations with a shift (SIES) have been studied for a long time (see [1,2]
and references therein). Many papers devoted to singular integral operators with a shift
(SIOS) are given to the construction of the Fredholm theory. Once M. G. Krein called the
Fredholm theory of linear operators a rough theory, and the theory describing its defect
subspaces a delicate theory [3]. However, the Fredholm theory of these operators brings
about only one thing, the defect of dimensions of kernel of the operator and its dual
operator. In other words, it is only the defect of the numbers of linear independent
solutions of the homogeneous equations reduced by the operator and the corresponding
dual operator. So the question of solving (and even of estimating the numbers of
solutions) of the corresponding equations actually remains open [4]. There are only a few
special types of SIES for which it is possible to answer this question to some extent [2,5].
Among the SIES of this type not reducible to two-term boundary value problems,
the most general and important is the class of singular integral equations with a
*Corresponding author. Email:
Complex Variables and Elliptic Equations
ISSN 1747-6933 print/ISSN 1747-6941 online ß 2008 Taylor & Francis
/>DOI: 10.1080/17476930701619782


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118

L. H. Chuan et al.

linear-fractional Carleman shift. In our view, the singular integral equations with a
linear-fractional Carleman shift in the unit circle, in addition, deserve the interest.
Factorization is the main method used by some authors to investigate Fredholm and
solvability theory for SIES (see [3,4] and references therein). In [6], two of us gave a
general formula of linear-fractional Carleman shifts on the unit circle and solved by

means of Riemann boundary value problem for a class of singular integral equations with
a linear-fractional Carleman shift on the unit circle. In this article, we study the
solvability for a class of singular integral equations with a linear-fractional Carleman
shift and with the degenerate kernels on the unit circle. In genaral, one knows that the
singular integral operator of Cauchy’s type (denoted by S) do not commute with
Carleman shift opetaor (denoted by W), but the difference beetwen them WS À SW is a
compact operator [2]). In section 2, we obtain some identities relating to those operators.
The scheme of our investigation is divided into two parts: first, we move the degenerate
kernels to the right-side hand of the equation. Based on the identity W n ¼ I, we construct
the orthogonal projectors and reduce the equation to a system of singular integral
equations without shift and solve this system by means of Riemann boundary value
problem. Second, we reconstruct the solution of the orginal equation from the solutions
of system of equations that can be solved, but its solution depends on some unknown
parameters. As indicated below, the equations of the type (1.1) can be solved by means of
Riemann boundary value problem and by of a system of linear algebraic equations.
Let À ¼ {t 2 C, jtj ¼ 1} be the unit circle on the complex plane C and let X :¼ H(À),
0551. Let
! ðtÞ ¼

t þ

,
t þ 

ð À
¼ 1;
6¼ 0Þ

be a Carleman linear-fractional function shift of order of n on À, i.e.,
8

! : À ! À;
>
>
>
<
!n ðtÞ ¼ t; for every t 2 À ð!k ðtÞ :¼ !ð!kÀ1 ðtÞÞ; !0 ðtÞ  tÞ;
>
for some t 2 À; k ¼ 1, 2, . . . , n À 1,
!
k ðtÞ 6 t;
>
>
:
! positive orientation of À
The such functions ! (t) might be of the form
8
tÀ
>
, if
n ¼ 2;
<
" À1
! ðtÞ ¼ t
>
: ei t À  , if n > 2,
" À1
t
" cos2 ðk=nÞ, for some k 2 {1, 2, . . . , n À 1},
where jj51; cos  ¼ 1 À 2ð1 À Þ
(k, n) ¼ 1 [6].

In this article, we study the solvability of the equations of the form
Z
nÀ1
bðtÞ X
’ ðÞ
nÀk 1
d
"
a ðtÞ ’ ðtÞ þ
n k¼0 ‘ i À  À !k ðtÞ
Z
m
X
1
þ
aj ðtÞbj ðÞ’ ðÞd ¼ fðtÞ,
i À
j¼1
where 1

‘

n À 1, "1 ¼ eð2i=nÞ, "‘ ¼ "‘1 :

ð1:1Þ


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Singular integral equations

119

2. Some identities of singular integral operator of Cauchy’s type and linear-fractional
shift operator on unit circle
Consider the following operators in X:
Z
1
’ ðÞ
d,
ðS’Þ ðtÞ ¼
i À  À t
ðW’Þ ðtÞ ¼ ’ð! ðtÞÞ,
n
1X
Pk ¼
"nÀ1Àj W jþ1 ,
n j¼1 k

ð2:1Þ
k ¼ 1, 2, . . . :

In the sequel, we shall need the following identities [7,8]:
8
n
P
>
k
>
W

¼
"kj Pj ; k ¼ 1; 2, . . . , n,
>
>
>
j¼1
>
<
Pk Pj ¼ kj Pj ; k; j ¼ 1; 2, . . . , n,
>
n
>
X
>
>
>
Pj ¼ I;
>
:

ð2:2Þ

j¼1

where kj is the Kronecker symbol. For every a 2 X we write (K a ’) (t) ¼ a (t) ’ (t).
Lemma 2.1 Let a 2 X be fixed. Then for every (k, j ), k, j 2 {1, 2, . . . , n} there exists
an element b 2 X such that Kb Xj & Xk and Pk Ka Pj ¼ Kb Pj , where xk :¼ P( X ).
The Function b(t) will be denoted by akj (t) and determined as follows
akj ðtÞ ¼

n
1X
" jÀk að!þ1 ðtÞÞ:
n ¼1 þ1

ð2:3Þ

By using (2.2) we get

Proof

Pk Ka Pj ¼

n
n
1X
1X
þ1
"nÀ1À
W
K
P
¼
"nÀ1À að!þ1 ðtÞÞW þ1 Pj
a
j
n ¼1 k
n ¼1 k

n

n
n
X
1X
1X
þ1
"Àðþ1Þ
að!
ðtÞÞ
"
P
P
¼
"Àðþ1Þ
að!þ1 ðtÞÞ"þ1
Pj
þ1

j

j
k
n ¼1 k
n
¼1
¼1
!
n
n
1X

1X
jÀk
Àk j
¼
" " að!þ1 ðtÞÞPj ¼
" að!þ1 ðtÞÞ Pj ¼ akj ðtÞPj ;
n ¼1 þ1 þ1
n ¼1 þ1

¼

where
ak j ðtÞ ¼

n
1X
" jÀk að!þ1 ðtÞÞ:
n ¼1 þ1

Putting akj(t) :¼ b(t), we obtain b 2 X and Pk Ka Pj ¼ Kb Pj.
Lemma 2.2

Let a 2 X be fixed. Then for any k, j 2 {1, 2, . . . , n}, we have
Pk Kak j ¼ Kak j Pj ,

ð2:4Þ
g


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120

L. H. Chuan et al.

where akj(t) are determined by (2.4).
Proof

For any ’ 2 X we have
n
1X
"nÀ1À Wþ1 akj ðtÞ’ðtÞ
n ¼1 k
!
!
n
n
1X
1X
Àðþ1Þ þ1
jÀk
"
W
" að!þ1 ðtÞÞ ’ðtÞ
¼
n ¼1 k
n ¼1 þ1
"
#
n

n
Á
1X
1X
Àðþ1Þ À
jÀk jÀk
"þ1 "þ1 að!þ1þþ1 ðtÞÞ "kÀj
W þ1 ’ ðtÞ
¼
þ1 "k
n ¼1 n ¼1

ðPk Kakj ’ÞðtÞ ¼ Pk akj ðtÞ’ðtÞ ¼

n
À þ1 Á
1X
Àk
akj ðtÞ"kÀj
’ ðtÞ
þ1 "þ1 W
n ¼1
"
#
n
1X
nÀ1À
þ1
"
W

’ðtÞ ¼ akj ðtÞðPj ’ÞðtÞ ¼ ðKakj Pj ÞðtÞ:
¼ akj ðtÞ
n ¼1 j

¼

g

Thus Pk Kakj  Kakj Pj :
Lemma 2.3

Let ’ 2 X. Then for every z 2 C we have

(1) ðSW ’ÞðzÞ ¼ ðW k S’ÞðzÞ À ðW kÀ1 S’Þ ð=
Þ; k ¼ 1; 2, . . . , W 0 ¼ I:
(2) ðPk S’ÞðzÞ ¼ ðSPk ’ÞðzÞ þ ð1="k À 1ÞðSPk ’Þ ð=
Þ; k ¼ 1; 2, . . . , n À 1,
k

where  and
are
!ðzÞ ¼ ððz þ
Þ=ð
z þ ÞÞ.

the

coefficients

of

the

linear-fractional

function

Proof
(1) By induction on k. For k ¼ 1 we have
Z
1
’ð!ðÞÞ
d:
ðSW’ÞðzÞ ¼
i À  À z
Put
 ¼ !À1 ðxÞ ¼

x À

;
À
x þ 

then d ¼

1
dx:
ðÀ
x þ Þ2

Therefore,
Z

’ðxÞð1=ð
x À Þ2 Þ
dx
À ðx À
= À
x þ Þ À z
Z
1
’ðxÞ
dx
¼
i À ðÀ
x þ Þðx À
À zðÀ
x þ ÞÞ

Z 
1
1
1
À
¼
’ðxÞdx
i À x À ðz þ
=
z þ Þ x À ð=

Þ
 
Z
Z
1
’ðxÞ
1
’ðxÞ

dx À
dx ¼ ðWS’ÞðzÞ À ðS’Þ
:
¼
i À x À !ðzÞ
i À x À ð=
Þ


ðSW’ÞðzÞ ¼

1
i


121

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Singular integral equations

So, we obtain
 

;
ðSW’ÞðzÞ ¼ ðWS’ÞðzÞ À ðS’Þ


for any z 2 C:

ð2:5Þ

Suppose that (i) is true for k ¼ m. For k ¼ m þ 1 we find
ðSW mþ1 ’ÞðzÞ ¼ ½SWðW m ’ފðzÞ
 
 


m
m
¼ W ½ðSW ’Þðzފ À ðSW ’Þ
¼ ½WSðW ’ފðzÞ À ½SðW ’ފ


 !
 
 !



À ðW m S’Þ

À ðW mÀ1 S’Þ
¼ W ðW m S’ÞðzÞ À ðW mÀ1 S’Þ



 !
 
 



À ðW m S’Þ
þ ðW mÀ1 S’Þ
:
¼ ðW mþ1 S’ÞðzÞ À W ðW mÀ1 S’Þ



m

m

Hence W ½ðW mÀ1 S’Þð=
ފ ¼ ðW mÀ1 S’Þð=
Þ; provided ðW mÀ1 S’Þð=
Þ is a constant.
Therefore
 

mþ1

mþ1
m
:
ðSW ’ÞðzÞ ¼ ðW S’ÞðzÞ À ðW S’Þ

The first part of the lemma is proved.
(2) Rewrite the equality in (1) in the form
ðW k S’ÞðzÞ ¼ ðSW k ’ÞðzÞ þ ðW kÀ1 S’Þ

 

:


We find
 !
n
n
1X
1X

nÀiÀ1
iþ1
nÀiÀ1
iþ1
i
"
ðW S’ÞðzÞ ¼
"
ðSW ’ÞðzÞ þ ðW S’Þ

ðPk S’ÞðzÞ ¼
n i¼1 k
n i¼1 k

"
! #
"
! # 
n
n
1X
1
1X

nÀiÀ1
iþ1
nÀi
i
"k
W
"k W S’
¼ S
’ ðzÞ þ
n i¼1
"k
n i¼1

 
1


¼ ðSPk ’ÞðzÞ þ ðPk S’Þ
:
"k

Substituting z ¼ ð=
Þ in formula (2.6), we get


 
 
1


ðPk S’Þ
¼ ðSPk ’Þ
:
1À
"k



ð2:6Þ

ð2:7Þ

If k 2 {1, 2, . . . , n À 1} then "k 6¼ 1. In this case, substituting (2.7) in (2.6) we receive
 
1

ðSPk Þ’

:
g
ðPk S’ÞðzÞ ¼ ðSPk ’ÞðzÞ þ
"k À 1



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122

L. H. Chuan et al.

Comment From the identity (2.5), one can say that the operators S and W do not
commute to each other, but the difference of WS and SW at the a function ’(t) always
equals ðS’Þð=
Þ:

3. Reducing equation (1.1) to a system of singular integral equations
We now represent the equation (1.1) in the following form
a ðtÞ’ ðtÞ þ bðtÞðP‘ S’ÞðtÞ þ

Z
m
X
1
aj ðtÞbj ðÞ’ðÞd ¼ f ðtÞ;
i À
j¼1

ð3:1Þ

where a, b, f, a1, . . . , am, b1, . . . , bm 2 X are given, and S, P‘ (1 ‘ n À 1) are the
operators defined by (2.1). Suppose that a(t) is a non-vanishing function on À.
Denote by Mbj ; j ¼ 1, . . . , m, the linear functionals on X defined as follows
Mbj ð’Þ ¼

1
i

Z
bj ðÞ’ðÞd;

for any ’ 2 X:

À

Put j ¼Mbj(’), j=1, . . . , m. We reduce equation (3.1) to the following problem:
Find solutions ’ of equation
aðtÞ’ðtÞ þ bðtÞðP‘ S’ÞðtÞ ¼ fðtÞ À

m
X

j aj ðtÞ

ð3:2Þ

j¼1

depended on the parameters 1, . . . , m, and fulfilled the following conditions
Mbj ð’Þ ¼ j ; j ¼ 1, . . . :

ð3:3Þ

Lemma 3.1 Let ’ 2 X. Then ’ is a solution of (3.2) if and only if {’k ¼ Pk’,
k ¼ 1, 2, . . . , n} is a solution of the following system
Ã

a ðtÞ ’k ðtÞ þ

bÃk‘ ðtÞðS’‘ ÞðtÞ

 
bÃk‘ ðtÞ

ðS’‘ Þ
þ
¼ fkà ðtÞ,

"‘ À 1

k ¼ 1, 2, . . . , n,

ð3:4Þ

where
aà ðtÞ ¼

n

Y

a ð! j þ1 ðtÞÞ;

j¼1

bÃk‘ ðtÞ ¼

n
n
Y
1X
jþ1
"‘Àk
bð!
ðtÞÞ
að! þ1 ðtÞÞ;
n j¼1 jþ1
¼1

ð3:5Þ

6¼j

f Ãk ðtÞ

"
#
n
m

n
Y
X
1X
nÀ1Àj
jþ1
jþ1
¼
"k
f ð! ðtÞÞ À
 a ð! ðtÞÞ
að! þ1 ðtÞÞ:
n j¼1
¼1
¼1
6¼j


123

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Singular integral equations

Proof

Suppose that ’ 2 X is a solution of (3.2). We then have
n
Y

að! þ1 ðtÞÞ’ðtÞ þ bðtÞ

¼1

"
¼ fðtÞ À

m
X

#
j aj ðtÞ

j¼1

n
Y

að!þ1 ðtÞÞðP‘ S’ÞðtÞ

¼1
6¼nÀ1

n
Y

að!þ1 ðtÞÞ:

¼1
6¼nÀ1

Applying the projections Pk, k ¼ 1, 2, . . . , n to both sides of above equation and using
the Lemmas 2.1 and 2.2, we obtain
2
3
n
n
X
Y
61
7
"‘Àk bð! jþ1 ðtÞÞ
að! þjþ2 ðtÞÞ5ðP‘ S’ÞðtÞ
aà ðtÞðPk ’ÞðtÞ þ 4
n j¼1 jþ1
¼1
6¼nÀ1

"

#
n
m
n
Y
X
1X
nÀjÀ1
jþ1
jþ1

¼
"k
f ð! ðtÞÞ À
 a ð! ðtÞÞ
að! þjþ2 ðtÞÞ:
n j¼1
¼1
¼1

ð3:6Þ

6¼nÀ1

It is easy to see that
n
Y

að! þjþ2 ðtÞÞ 

¼1
6¼nÀ1

n
Y

að!þ1 ðtÞÞ

for any j 2 f1; 2, . . . , ng:

¼1

6¼j

Hence, (3.6) is equivalent to the following system
aà ðtÞðPk ’ÞðtÞ þ bÃk‘ ðtÞðP‘ S’ÞðtÞ ¼ f Ãk ðtÞ; k ¼ 1; 2, . . . , n:

ð3:7Þ

Using Lemma 2.3, we rewrite the system (3.7) in the form
 
bà ðtÞ

aà ðtÞðPk ’ÞðtÞ þ bÃk‘ ðtÞðSP‘ ’ÞðtÞ þ k‘ ðSP‘ ’Þ
¼ f Ãk ðtÞ, k ¼ 1, 2, . . . , n:

"‘ À 1
Thus (P1’, P2’ , . . . , Pn’) is a solution of (3.4).
Conversely, suppose that there exists  2 X such that (P1’, P2’ , . . . , Pn’) is a solution
of (3.4). Summing by k from 1 to n, we obtain
 ! X
n
n
X
1

Ã
Ã
ðSP‘ ’Þ
bk‘ ðtÞ ðSP‘ ’ÞðtÞ þ
f Ãk ðtÞ:
ð3:8Þ

¼
a ðtÞ ’ ðtÞ þ
"

À
1
‘
k¼1
k¼1
From (3.5), we get
n
X
k¼1

n
n
X
1X

jþ1
"‘Àk
ðtÞÞ
jþ1 bð!

n
Y

að! þ1 ðtÞÞ
n
¼1

j¼1
k¼1
6¼j
"
#
n
n
n
X 1X
Y
jþ1
¼
"‘Àk
ðtÞÞ
að! þ1 ðtÞÞ
bð!
jþ1
n
¼1
j¼1
k¼1

bÃk‘ ðtÞ ¼

6¼j

¼ bðtÞ

n
Y

¼1
6¼nÀ1

að! þ1 ðtÞÞ:

ð3:9Þ


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124

L. H. Chuan et al.

Similarly,
n
X

"
f

Ã
k ðtÞ

¼

f ðtÞ À

m
X

#

j¼1

k¼1

n
Y

j aj ðtÞ

að! þ1 ðtÞÞ:

ð3:10Þ

¼1
6¼nÀ1

Therefore, (3.8) is equivalent to the following equality
aà ðtÞ’ðtÞ þ bðtÞ

n
Y

að! þ1 ðtÞÞ ðSP‘ ’ÞðtÞ þ

¼1
6¼nÀ1

¼ fðtÞ À

m
X
j¼1

 !
1

ðSP‘ ’Þ
"‘ À 1


!Y
n
j aj ðtÞ
að! þ1 ðtÞÞ:
¼1
6¼nÀ1

This implies
aðtÞ’ðtÞ þ bðtÞðP‘ S’ÞðtÞ ¼ fðtÞ À

m
X

g

j aj ðtÞ:

j¼1

Lemma 3.2 If (’1, ’2, . . . , ’n) is a solution of system (3.4) then (P1’1, P2’2 , . . . , Pn’n)
is also its solution.
Proof Suppose (1, 2, . . . ,n) is a solution of the system (3.4). Applying the
projections Pk to both sides of k-th equation of (3.4) we get
aà ðtÞðPk ’k ÞðtÞ þ Pk bÃk‘ ðtÞðS’‘ ÞðtÞ þ

 !
bÃk‘ ðtÞ

ðS’‘ Þ
¼ Pk ðf Ãk ðtÞÞ:

"‘ À 1

ð3:11Þ

It is easy to see that
Pk ðf Ãk ðtÞÞ ¼ f Ãk ðtÞ and

Pk bÃk‘ ðtÞ ¼ bÃk‘ ðtÞP‘ :

ð3:12Þ

Substituting (3.12) into (3.11), we obtain
Ã

a ðtÞðPk ’k ÞðtÞ þ

bÃk‘ ðtÞðP‘ S’‘ ÞðtÞ


 
bÃk‘ ðtÞ

P‘ ðS’‘ Þ
þ
¼ f Ãk ðtÞ:

"‘ À 1

ð3:13Þ

Provided ðS’‘ Þð=
Þ is a constant function, then

P‘

 

 
n

1X

nÀjÀ1
jþ1
ðS’‘ Þ
"

W
ðS’‘ Þ
¼

n j¼1 ‘

 X
1
 n nÀjÀ1
¼ ðS’‘ Þ
"
¼ 0:
n
j¼1 ‘

ð3:14Þ

Using Lemma 2.3, (3.13) is equivalent to the following equation
aà ðtÞðPk ’k ÞðtÞ þ bÃk‘ ðtÞðSP‘ ’‘ ÞðtÞ þ

 
bÃk‘ ðtÞ

ðSP‘ ’‘ Þ
¼ f Ãk ðtÞ;

"‘ À 1

Thus, (P1’1, P2’2, . . . , Pn’n) is a solution of (3.4).

k ¼ 1; 2, . . . , n:


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Singular integral equations

Theorem 3.1

125

The equation (3.2) has solutions in X if and only if the following equation
aà ðtÞ’‘ ðtÞ þ bÑ‘ ðtÞðS’‘ ÞðtÞ þ

 
bÑ‘ ðtÞ

ðS’‘ Þ
¼ f Ñ ðtÞ
"‘ À 1


ð3:15Þ

has solutions. Moreover, if ’‘(t) is a solution of equation (3.15) then equation (3.2) has a
solution given by formula
’ðtÞ ¼

f ðtÞ À

Pm

j¼1

j aj ðtÞ À bðtÞðP‘ S’‘ ÞðtÞ
:
aðtÞ

ð3:16Þ

Proof Suppose that ’ 2 X is a solution of equation (3.2). By Lemma 3.1,
(P1’, P2’, . . . , Pn’) is a solution of system (3.4). Hence, P‘’ is a solution of (3.15).
Conversely, suppose that ’‘(t) is a solution of (3.15). In this case, system (3.4) has
solution (’1, ’2, . . . , ’n) determined by the formula
f Ãk ðtÞ À bÃk‘ ðtÞðS’‘ ÞðtÞ À ððbÃk‘ ðtÞÞ=ð"‘ À 1ÞÞðS’‘ Þð=
Þ
;
aà ðtÞ
k ¼ 1, 2, . . . , n; k 6¼ ‘:

’k ðtÞ ¼

ð3:17Þ

By Lemma 3.2, we have (P1’1, P2’2, . . . , Pn’n) is also a solution of (3.4). Put
’¼

n
X

Pk ’k :

ð3:18Þ

k¼1

It is clear that Pk’ ¼ Pk’k. This means that (P1’, P2’ , . . . , Pn’) is a solution of (3.4).
From Lemma 3.1 it follows that  is a solution of (3.2). Moreover, from (3.17) and
(3.18) we get
’ðtÞ ¼

n
X
k¼1

¼

Pk ’k ¼

n
X
k¼1

Pk

f Ãk ðtÞ À bÃk‘ ðtÞðS’‘ ÞðtÞ À ðbÃk‘ ðtÞÞ=ð"‘ À 1ÞðS’‘ Þð=
Þ
aà ðtÞ



 !
n
bÃk‘ ðtÞ
1 X

Ã
Ã
P
f
ðtÞ
À
b
ðtÞðP
S’
ÞðtÞ
À
ðS’
Þ
:
‘
‘
‘
‘
k‘
aà ðtÞ k¼1 k

"‘ À 1

ð3:19Þ

Substituting (3.9), (3.10), (3.14) into (3.19) we obtain
P
f ðtÞ À m
j¼1 j aj ðtÞ À bðtÞðP‘ S’‘ ÞðtÞ
’ðtÞ ¼
:
aðtÞ
g

The proof is complete.
4. The solvability of equation (3.15)
We set
Dþ ¼ fz 2 C : jzj51g;

DÀ ¼ fz 2 C : jzj41g


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126

L. H. Chuan et al.

Denote by H(Dþ),H(DÀ) the sets of the analytic functions in Dþ and DÀ respectively.
Consider the equation (3.15)
 
bà ðtÞ

aà ðtÞ’‘ ðtÞ þ bÑ‘ ðtÞðS’‘ ÞðtÞ þ ‘‘ ðS’‘ Þ
¼ f Ñ ðtÞ:

"‘ À 1

Put
È‘ ðzÞ ¼

1
2i

Z

’‘ ðÞ
d;
ÀÀz

z 2 CnÀ:

According to Sokhotski–Plemelij formula, we have [1]
À
’‘ ðtÞ ¼ Èþ
‘ ðtÞ À È‘ ðtÞ,

ð4:1Þ

À
ðS’‘ ÞðtÞ ¼ Èþ
‘ ðtÞ þ È‘ ðtÞ:

ð4:2Þ

Moreover, ðS’‘ Þð=

Þ ¼ 2È‘ ð=
Þ: Put 0 ¼ È‘ ð=
Þ. We reduce equation (3.15) to the
following boundary problem: find a sectionally analytic function È‘(z) (È(z) ¼ Èþ(z)
for z 2 Dþ, È(z) ¼ ÈÀ(z) for z 2 DÀ) vanishes at infinity, È‘ ð=
Þ ¼ 0 and satisfies the
following linear relation on À
À
Èþ
‘ ðtÞ ¼ GðtÞÈ‘ ðtÞ þ gðtÞ;

t 2 À;

ð4:3Þ

where
GðtÞ ¼

aà ðtÞ À bÑ‘ ðtÞ
f à ðtÞ À 0 ðð2bÑ‘ ðtÞÞ=ð"‘ À 1ÞÞ
; gðtÞ ¼ ‘
:
Ã
Ã
a ðtÞ þ b‘‘ ðtÞ
aà ðtÞ þ bÑ‘ ðtÞ

ð4:4Þ

Suppose that aà ðtÞ Æ bÑ‘ ðtÞ are the non-vanishing functions on À. Then G(t), g(t) 2 X and

G(t) 6¼ 0 for any t 2 À. Put
Z
1
d ln GðtÞ,
{ ¼ GðtÞ ¼
2i À
Z
1
ln½ À{ Gðފ
d,
ÀðzÞ ¼
2i À
Àz
þ

Xþ ðzÞ ¼ eÀ

ðzÞ

,

À

XÀ ðzÞ ¼ zÀ{ eÀ

ðzÞ

:

ð4:5Þ

Using the results in [2] (p. 16–20) we get the following cases:
(1) { ! 0. The equation (4.3) has general solution is given by formula
!
Z
1
gðÞ d
È‘ ðzÞ ¼ XðzÞ
þ P{À1 ðzÞ
2i À Xþ ðÞ  À z
Â
Ã
¼ XðzÞ ÉðzÞ À 0 BðzÞ þ P{À1 ðzÞ ,

ð4:6Þ

where
ÉðzÞ ¼

1
2i

Z

f‘Ã ðÞ
d
,
Ã
þ
Ã

À X ðÞða ðÞ þ b‘‘ ðÞÞ  À z

ð4:7Þ


127

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Singular integral equations

1
BðzÞ ¼
2i

Z

ð2bÑ‘ ðÞ="‘ À 1Þ
d
,
þ ðÞðaà ðÞ þ bà ðÞÞ  À z
X
À
‘‘

P{À1 ðzÞ  0,

ð4:8Þ

if { ¼ 0,

and
P{À1 ðzÞ ¼ p1 þ p2 z þ Á Á Á þ p{ z{À1 ,

if { ! 1,

ð4:9Þ

which is a polynomial of degree { À 1 with arbitrary complex coefficients. The
function È‘(z) determined in (4.6) is a solution of problem (4.3) if È‘ ð=
Þ ¼ 0 ,
that is
   
 
 !




É
À 0 B
þ P{À1
¼ 0 :
X




This implies
   !

   
 !





0 1 þ X
B
¼X
É
þ P{À1
:






ð4:10Þ

(i) If 1 þ Xð=
ÞBð=
Þ 6¼ 0 : from (4.10), we get
0 ¼

Xð=
Þ½Éð=
Þ þ P{À1 ð=
ފ

:
1 þ Xð=
ÞBð=
Þ

ð4:11Þ

In this case, the general solution of problem (4.3) is given by formula
!
Xð=
Þ½Éð=
Þ þ P{À1 ð=
ފ
BðzÞ þ P{À1 ðzÞ ,
È‘ ðzÞ ¼ XðzÞ ÉðzÞ À
1 þ Xð=
ÞBð=
Þ

ð4:12Þ

where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8) and P{À1(z) is a
polynomial of degree { À 1 with arbitrary complex coefficients.
(ii) If 1 þ Xð=
ÞBð=
Þ ¼ 0 : from (4.10) we get
 
 



þ P{À1
¼ 0:
ð4:13Þ
É


Then, the general solution of problem (4.3) is given by formula
È‘ ðzÞ ¼ XðzÞ½ÉðzÞ À 0 BðzÞ þ P{À1 ðzފ,

ð4:14Þ

where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8), 0 is arbitrary and
P{À1(z) is a polynomial of degree { À 1 with complex coefficients satisfying
condition (4.13).
(2) {50. The necessary condition for the problem (4.3) to be solvable is that
Z
gðÞ {À1
 d ¼ 0, k ¼ 1, . . . , À{:
þ
À X ðÞ


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128

L. H. Chuan et al.

This condition can be written as follows
Z

f Ñ ðÞ {À1
d ¼ 0
Ã
þ
Ã
À X ðÞða ðÞ þ b‘‘ ðÞÞ

Z

ðð2bÑ‘ ðÞÞ=ð"‘ À 1ÞÞ kÀ1
d;
Ã
þ
Ã
À X ðÞða ðÞ þ b‘‘ ðÞÞ

k ¼ 1, . . . , À{: ð4:15Þ

In this case, we have P{À1(z)  0. So we receive
(i) If 1 þ Xð=
ÞBð=
Þ 6¼ 0 : from (4.11) we get
0 ¼

Xð=
ÞÉð=
Þ
:
1 þ Xð=

ÞBð=
Þ

Hence, (4.15) becomes the following condition
Z

f Ñ ðÞ kÀ1
Xð=
ÞÉð=
Þ
d ¼
þ ðÞðaà ðÞ þ bà ðÞÞ
1
þ
Xð=
ÞBð=
Þ
X
À
‘‘

Z

ð2bÑ‘ ðÞ="‘ À 1Þ kÀ1
d;
Ã
þ
Ã
À X ðÞða ðÞ þ b‘‘ ðÞÞ

ð4:16Þ

k ¼ 1, . . . , À {:
If the condition (4.16) is satisfied then the solution of the problem (4.3) is given
by formula
!
Xð=
ÞÉð=
Þ
BðzÞ ;
È‘ ðzÞ ¼ XðzÞ ÉðzÞ À
1 þ Xð=
ÞBð=
Þ

ð4:17Þ

where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8).
(ii) If 1 þ Xð=
ÞBð=
Þ ¼ 0 : from (4.13) we get
É

 

¼ 0:


ð4:18Þ

If the conditions (4.15) and (4.18) are satisfied then the solution of the problem
(4.3) is given by formula
Â
Ã
È‘ ðzÞ ¼ XðzÞ ÉðzÞ À 0 BðzÞ ;

ð4:19Þ

where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8) and 0 is determined
from condition (4.15).
Now we can formulate the main results about solutions of the equation (3.15)
in the following form
Theorem 4.1

Suppose that the functions aà ðtÞ Æ bÑ‘ ðtÞ does not vanish on À.

(1) If 1 þ Xð=
ÞBð=
Þ 6¼ 0 and { ! 0 then equation (3.15) has solutions ’‘ which
satisfy the following formula
!
Xð=
Þ½Éð=
Þ þ P{À1 ð=
ފ þ
B ðtÞ þ P{À1 ðtÞ
S’‘ ðtÞ ¼ Xþ ðtÞ Éþ ðtÞ À
1 þ Xð=
ÞBð=
Þ

!
Xð=
Þ½Éð=
Þ
þ P{À1 ð=
ފ þ X À ðtÞ ÉÀ ðtÞ À
B ðtÞ þ P{À1 ðtÞ ,
ð4:20Þ
1 þ Xð=
ÞBð=
Þ


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Singular integral equations

129

where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8) and P{ À 1(z) is a
polynomial of degree { À 1 with arbitrary complex coefficients.
(2) If 1 þ Xð=
ÞBð=
Þ 6¼ 0 and {50 then equation (3.15) is solvable if the condition
(4.16) is satisfied. In this case, equation (3.15) has unique solution which satisfies the
formula (4.20), where P{À1(z)  0.
(3) If 1 þ X=
ÞBð=
Þ ¼ 0 and { ! 0 then equation (3.15) has solutions ’‘ which

satisfy the following formula
Â
Ã
S’‘ ðtÞ ¼ Xþ ðtÞ Éþ ðtÞ À 0 Bþ ðtÞ þ P{À1 ðtÞ
þ XÀ ðtÞ½ÉÀ ðtÞ À 0 BÀ ðtÞ þ P{À1 ðtފ,

ð4:21Þ

where X(z), É(z), B(z) are determined by (4.5), (4.7), (4.8), 0 is arbitrary and
P{À1(z) is a polynomial of degree { À 1 with complex coefficients satisfying the
condition (4.13).
(4) If 1 þ Xð=
ÞBð=
Þ ¼ 0 and {50 then the equation (3.15) is solvable if the
condition (4.15) and (4.18) are satisfied. In this case, equation (3.15) has unique
solution which satisfies the formula (4.21), where P{À1(z)  0 and 0 is determined
from the condition (4.15).
Proof (1) From assumption it follows that the problem (4.3) has a solution È‘(z)
determined by (4.12). Therefore, equation (3.15) has a solution ’‘(t) determined by
(4.1). Moreover, from (4.2) we get
À
S’‘ ðtÞ ¼ Èþ
‘ ðtÞ þ È‘ ðtÞ

!
Xð=
Þ ½Éð=
Þ þ P{À1 ð=
ފ þ
B ðtÞ þ P{À1 ðtÞ

1 þ Xð=
ÞBð=
Þ
!
Xð=
Þ
½Éð=
Þ þ P{À1 ð=
ފ þ XÀ ðtÞ ÉÀ ðtÞ À
B ðtÞ þ P{À1 ðtÞ :
1 þ Xð=
ÞBð=
Þ

¼ Xþ ðtÞ Éþ ðtÞ À

Similarly, the cases (2), (3), (4) can be proved.

g

5. The solvability of equation (3.1)
Theorems 3.1 and 4.1 show that if aà ðtÞ Æ bÑ‘ ðtÞ 6¼ 0 on À then equation (3.2) is solvable
in a closed form. In this section, we study which solutions of (3.2) will be the solution of
(3.1), i.e., the solutions of (3.2) need to satisfy the condition (3.3). Consider the
following cases:
(1) 1 þ Xð=
ÞBð=
Þ 6¼ 0; { ! 0: By using Theorems 3.1 and 4.1, we have solutions
of (3.2) given by the following formula

’ ðtÞ ¼

f ðtÞ À

Pm

j¼1

j aj ðtÞ À bðtÞðP‘ S’‘ ÞðtÞ
;
aðtÞ

ð5:1Þ


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130

L. H. Chuan et al.

where S’‘(t) is determined by (4.20). From (3.5) and (4.7) we get
Â
Ã
Z ð1=nÞ Pn "nÀ1Àj f ð! jþ1 ðÞÞ À Pm  a ð!jþ1 ðÞÞ Qn¼1 að! þ1 ðÞÞ
j¼1 ‘
¼1
1
d
6¼j

ÉðzÞ ¼
Ã
þ
Ã
2i À
Àz
X ðÞða ðÞ þ b‘‘ ðÞÞ
m
X
¼ É1 ðzÞ À
 A ðzÞ,
ð5:2Þ
¼1

where
Z ð1=nÞ Pn "nÀ1Àj f ð! jþ1 ðÞÞ Qn¼1 að!þ1 ðÞÞ
j¼1 ‘
d
6¼j
,
Ã
þ
Ã
Àz
X ðÞða ðÞ þ b‘‘ ðÞÞ
À
Z ð1=nÞ Pn "nÀ1Àj a ð! jþ1 ðÞÞ Qn¼1 að! þ1 ðÞÞ
j¼1 ‘
1
d

6¼j
:
A ðzÞ ¼
2i À
Àz
Xþ ðÞðaà ðÞ þ bÑ‘ ðÞÞ

1
É1 ðzÞ ¼
2i

ð5:3Þ

ð5:4Þ

Substituting (4.9), (5.2) into (4.20) we find
À
À
S’‘ ðtÞ ¼ ½Xþ ðtÞÉþ
1 ðtÞ þ X ðtÞÉ1 ðtފ À

m
X

À
À
j ½Xþ ðtÞAþ
j ðtÞ þ X ðtÞAj ðtފ

j¼1

Â

Pm

Xð=
Þ É1 ð=
Þ À j¼1 j Aj ð=
Þ þ
À
1 þ Xð=
ÞBð=
Þ
þ

{
X

P{

j¼1 pj ð=
Þ

jÀ1

Ã
½X þ ðtÞB þ ðtÞ þ X À ðtÞB À ðtފ

pj t jÀ1 ½X þ ðtÞ þ X À ðtފ:

j¼1

Since we can rewrite (5.1) in the form
’ðtÞ ¼

f ðtÞ À b ðtÞP‘ ½X þ ðtÞÉ1 þ ðtÞ þ X À ðtÞÉÀ
1 ðtފ
aðtÞ
m
X

À
À
aj ðtÞ À bðtÞP‘ ½Xþ ðtÞAþ
j ðtÞ þ X ðtÞAj ðtފ
aðtÞ
j¼1
Â
Ã
P
P{
jÀ1
Xð=
Þ É1 ð=
Þ À m
j¼1 j Aj ð=
Þ þ
j¼1 pj ð=
Þ
þ

1 þ Xð=
ÞBð=
Þ

À

j

bðtÞP‘ ½X þ ðtÞB þ ðtÞ þ X À ðtÞB À ðtފ
aðtÞ
Ã
 jÀ1 þ
{
X bðtÞP‘ t ½X ðtÞ þ XÀ ðtފ
,
À
pj
aðtÞ
j¼1

Â

ð5:5Þ

where X(z), B(z), É1(z), A1(z), . . . , Am(z) are determined by (4.5), (4.8), (5.3), (5.4), and
p1, . . . , p{ are arbitrary. The function ’ is a solution of the equation (3.1) if it satisfies
the condition (3.3), that is
Mbk ’ ¼ k ;

k ¼ 1, . . . , m:

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Singular integral equations

131

Substituting (5.5) into the last condition, we obtain
"  
  X
 jÀ1 #
m
m
{
{
X
X
X



k ¼ dk À
ekj j þ É1
j Aj
pj
gkj pj
fk À
À
þ

j¼1
j¼1
j¼1
j¼1
"
  ! X
 !
 jÀ1 #
m
{
X



¼ dk þ É1
ekj þ fk Aj
gkj À fk
pj ;
fk À
j À



j¼1
j¼1
k ¼ 1, 2, . . . , m,

ð5:6Þ

where

dk ¼ Mbk
ekj ¼ Mbk


À
À
f ðtÞ À bðtÞP‘ ½Xþ ðtÞÉþ
1 ðtÞ þ X ðtÞÉ1 ðtފ
;
aðtÞ
!
À
À
aj ðtÞ À bðtÞP‘ ½Xþ ðtÞAþ
j ðtÞ þ X ðtÞAj ðtފ
;
aðtÞ




Xð=
Þ
bðtÞP‘ ½X þ ðtÞB þ ðtÞ þ X À ðtÞBÀ ðtފ
;
fk ¼ Mbk

1 þ Xð=
ÞBð=
Þ
aðtÞ
Ã
Â


bðtÞP‘ tjÀ1 ½Xþ ðtÞ þ XÀ ðtފ
gkj ¼ Mbk
:
aðtÞ

ð5:7Þ

Put
  1

0 1
B d1 þ É 1
f 1 C
p1
1
C
B
C
B
B . C
B .. C
.

C ,
B
B
C
.
¼@ . A , P¼@ . A , D¼B
..
C
  C
B
A
@

m mÂ1
p{ {Â1
dm þ É 1
fm

mÂ1
0
 
  1


B e11 þ f1 A1
Á Á Á e1m þ f1 Am
C
C
B
C

B
.. C
..
..
,
E¼B
C
B .
  .
.  C
B
@

 A
em1 þ fm A1
Á Á Á emm þ fm Am


mÂm
0
 0
 {À1 1


C
B g11 À f1
Á Á Á g1{ À f1
C
B
C

B
.
.
.
C
B
G¼B
..
..
..
C
C
B




@
 0
 {À1 A
gm1 À fm
Á Á Á gm{ À fm


mÂ{:
0

1

0

ð5:8Þ

Now we write (5.6) in the form of matrix condition
ðI þ EÞ ¼ D À GP,

ð5:9Þ

where I is the unit matrix. So we can formulate that the function ’ determined by (5.5) is
a solution of (3.1) if and only if (1 , . . . , m) satisfy the condition (5.9).


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132

L. H. Chuan et al.

(2) 1 þ Xð=
ÞBð=
Þ 6¼ 0; {50: From Theorems 3.1 and 4.1 it follows that the
equation (3.2) has solutions if and only if the condition (4.16) satisfied. If this is in
case, then P{À1  0. So, the solutions of (3.2) are given by as follows
’ðtÞ ¼

À
À
fðtÞ À bðtÞP‘ ½Xþ ðtÞÉþ
1 ðtÞ þ X ðtÞÉ1 ðtފ
aðtÞ

À
À
m
X
aj ðtÞ À bðtÞP‘ ½Xþ ðtÞAþ
j ðtÞ þ X ðtÞAj ðtފ
À
j
aðtÞ
j¼1
Â
Ã
Pm
Xð=
Þ É1 ð=
Þ À j¼1 j Aj ð=
Þ bðtÞP‘ ½Xþ ðtÞBþ ðtÞ þ XÀ ðtÞBÀ ðtފ
:
þ
aðtÞ
1 þ Xð=
ÞBð=
Þ

ð5:10Þ

Therefore, the function  determined by (5.10) is a solution of the equation (3.1) if and
only if (1, . . . , m) satisfy the following matrix condition
ðI þ E Þ ¼ D,

ð5:11Þ

where E and D are determined by (5.8). On the other hand, substituting (3.5), (5.2) into
(4.16) we get
 
 
m
m
X
 0 X

0
0
0
dk À
ek  ¼ É1
fk Aj
fk À
j , k ¼ 1, 2, . . . , À{ ,
ð5:12Þ


¼1
j¼1
where
Z ð1=nÞ Pn "nÀ1Àj f ð! jþ1 ðÞÞ Qn¼1 að! þ1 ðÞÞ
j¼1 ‘

6¼j
 kÀ1 d

þ ðÞðaà ðÞ þ bà ðÞÞ
X
À
‘‘
Z ð1=nÞ Pn "nÀ1Àj a ð! jþ1 ðÞÞ Qn¼1 að! þ1 ðÞÞ
j¼1 ‘
6¼j
e0k ¼
 kÀ1 d
Xþ ðÞðaà ðÞ þ bÑ‘ ðÞÞ
À
Z
Xð=
Þ
ðð2bÑ‘ ðÞÞ=ð"‘ À 1ÞÞ kÀ1
0
 d:
fk ¼
1 þ Xð=
ÞBð=
Þ À Xþ ðÞðaà ðÞ þ bÑ‘ ðÞÞ

dk0

¼

ð5:13Þ

Put
 

1
 0
0
À
É
f
d
1
B 1
1 C
C
B
C
B
C
B
0
.
D ¼B
C
..
C
B
C
B
 
A
@

0

0
dÀ{ À É1
fÀ{

À{Â1,
 
 
1
0


0
0
0
0
À
f
A
À
f
A
Á
Á
Á
e
e
1 1
1m
1 m
C

B 11


C
B
C
B
C
B
0
.
.
.
E ¼B
:
C
..
..
..
C
B
B
 
 C
@

 A
0
0
0

0
eÀ{1
À fÀ{
A1
À fÀ{
Am
Á Á Á eÀ{m


À{Âm
0

ð5:14Þ


133

Downloaded By: [Tuan, Nguyen Minh] At: 18:21 23 January 2008

Singular integral equations

We write (5.12) in the form of matrix condition
E 0 ¼ D 0:

ð5:15Þ

Combining (5.11) and (5.15) we can say that the function  determined by
(5.10) is a solution of (3.1) if and only if (1 , . . . , m) satisfy the following matrix
condition



 0
D
IþE

¼
:
ð5:16Þ
0
D ðmÀ{ÞÂ1
E
ðmÀ{ÞÂm
(3) 1 þ Xð=
ÞBð=
Þ ¼ 0, { ! 0: Then the solutions of the equation (3.2) are given by
the following formula
’ðtÞ ¼

À
À
fðtÞ À bðtÞP‘ ½Xþ ðtÞÉþ
1 ðtÞ þ X ðtÞÉ1 ðtފ
aðtÞ
À
À
m
X
aj ðtÞ À bðtÞP‘ ½Xþ ðtÞAþ
j ðtÞ þ X ðtÞAj ðtފ
À

j
aðtÞ
j¼1

bðtÞP‘ ½Xþ ðtÞBþ ðtÞ þ XÀ ðtÞBÀ ðtފ
aðtÞ
Ã
 jÀ1 þ
{
X
bðtÞP‘ t ½X ðtÞ þ XÀ ðtފ
,
À
pj
aðtÞ
j¼1

þ 0

ð5:17Þ

where X(z), B(z), É1(z), A1(z), . . . , Am(z) are determined by (4.5), (4.8), (5.3), (5.4), 0 is
an arbitrary complex number and p1, . . . , p{ satisfy the condition (4.13). Substituting
(5.2) in (4.13) we obtain
  X
  X
 jÀ1
m
{




É1
j Aj
pj
¼ 0:
À
þ
ð5:18Þ



j¼1
j¼1
The function ’ is a solution of the equation (3.1) if it satisfies the condition (3.3).
Substituting (5.17) into (3.3) we get
k ¼ dk À

m
X

ekj j þ 0 fk À

j¼1

¼ dk þ 0 fk À

{
X

gkj pj

j¼1
m
X

ekj j À

j¼1

{
X

gkj pj ,

k ¼ 1, 2, . . . , m,

where dk, ekj, fk, gkj are determined by (5.7). Put
0 1
0 1
0 1
f1
1
d1
B . C
B .. C
B .. C
C
¼@ . A ,
D¼@ . A ,

F¼B
@ .. A ,
m mÂ1
dm mÂ1
fm mÂ1
0
1
0
1
g11 Á Á Á g1m
e11 Á Á Á e1m
B.
.. C
..
. . .. C
B.
C
E ¼ @ ..
,
G¼B
. A
.
. . A
@ ..
em1

ÁÁÁ

emm

mÂm

ð5:19Þ

j¼1

gmj

Á Á Á gm{

mÂ{

0

1
p1
B . C
C
P¼B
@ .. A ,
p{ {Â1
:

ð5:20Þ