Some Properties of a Kind of Singular Integral Operator with Weierstrass Function Kernel ()
1. Introduction
The properties of singular integral operator with Cauchy or Hilbert kernel on simple closed smooth curve or open arc have been elaborately discussed in [1-3]. Based on these, for the boundary curve is a closed curve or an open arc, the authors discussed the singular integral operators and corresponding equation with Cauchy kernel or Hilbert kernel in [1-3]. In recent years, many authors discussed the numerical solution of a class of systems of Cauchy singular integral equations with constant coefficients, Numerical methods for nonlinear singular Volterra integral equations in [4-6].
In this paper, we consider a kind of singular integral operator with Weierstrass function kernel on a simple closed smooth curve in a fundamental period parallelogram. Our goal is to develop the Bertrand poincaré formula for changing order of the corresponding integration, and some important properties of the above singular integral operator.
2. Preliminaries
Definition 1 Suppose that
are complex constants with
, and P denotes the fundamental period parallelogram with vertices
. Then the function

is called the Weierstrass
-function, where

denotes the sum of all
, except for
.
Definition 2 Suppose that
is a smooth closed curve in the counterclockwise direction, lying entirely in the fundamental period parallelogram P, with
and the origin lying in the domain
enclosed by
. The following operator
(1)
is called the singular integral operator with
-function kernel on
, where
is the unknown function, and

are the given functions.
Letting
, then (1) becomes
(2)
Since
is uniformly convergent in any closed bounded region lying entirely in P,

for any
, where
is some positive finite constant. By noting that
, we obtain

, where
is some positive finite constant. Write



then (1) can be rewritten in the form
, (3)
where
is a Fredholm operator and
is called the characteristic operator of
. Now the index of
is defined as
, where

and for definiteness we assume that
, namely we assume that
is an operator of normal type.
Now the associated operator of (1) takes the form
(4)
or
(4)′
and so that the associated operator of
becomes

In addition, if we write

then (4) can be rewritten as
(5)
where
(
,
is some finite constant).
So
is a Fredholm operator, and then the characteristic operator of
operator becomes
(6)
Therefore, we concluded that
usually can not be established, that is
.
For convenience, we write

where the fixed nonzero point
and the origin lie in
. It is not difficult to get the following results.
Lemma 1 Suppose that
, and with the same
as mentioned before, then a) 

b) (Poincare-Bertrand formula)

3. Some Properties of Operator K
1) If
, then
.
Proof Through calculation and estimation, we have
(7)
for any
, where
and
are all finite constant. While for any
, we have
(8)
where
is some finite constant. Substituting (8) into (7), we obtain
(9)
Similarly we know that

Consequently, we have
.
2) If
are singular integral operator, then
is also a singular integral operator. That is, if

then
, (10)
where the sum of the former two terms in the right hand of Equation (10) are the characteristic operator, and the remainder in that is a Fredholm operator.
Proof By definition, we deduce that

where

By virtue of Lemma 1 (b),
can be rewritten in the form

Consequently, (10) is established.
Now we write

where
,
,
,
.
By [1], we know that
is a Fredholm integral. For
, we know from

that
is continuous about the variable
, and so that
is also a Fredholm integral. By nothing that
have the same form, we only need to discuss either one of them. Here we consider the integral
. Write

then
is analytic in P and so that
. Consequently, we read from

that
and so that
is continuous on
, therefore
is also a Fredholm integral.
So far, we conclude that
is a singular integral operator.
3) Let
, where
denotes the indices of
, then
.
Proof From 2), we know

and

so
.
In addition, we can see from
and
that when
are normal,
is also normal.
4)
.
5) If
is a singular integral operator, and
is a Fredholm integral operator of the first kind, then
and
are also Fredholm integral operators of the first kind.
6) If the indies of
and
are
and
respectively , then
.
7)
.
Through careful calculation, we may obtain 4) - 7).
8) Generally speaking,

can not be established for
.
Proof By definition and calculation, we have
. (11)
Whereas
. (12)
Let

then by Lemma 1(a), we have
, (13)
Substituting (13) into (12), we see that
. (14)
Therefore,
cannot be established.