Applied Mathematics, 2013, 4, 1231-1235 http://dx.doi.org/10.4236/am.2013.48166 Published Online August 2013 (http://www.scirp.org/journal/am)
Some Properties of a Kind of Singular Integral Operator with Weierstrass Function Kernel Lixia Cao Department of Information and Computing Sciences, Mathematics College, Northeast Petroleum University, Daqing, China Email:
[email protected] Received June 16, 2013; revised July 16, 2013; accepted July 25, 2013 Copyright © 2013 Lixia Cao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
ABSTRACT We considered a kind of singular integral operator with Weierstrass function kernel on a simple closed smooth curve in a fundamental period parallelogram. Using the method of complex functions, we established the Bertrand Poincaré formula for changing order of the corresponding integration, and some important properties for this kind of singular integral operator. Keywords: Weierstrass Function Kernel; Singular Integral Operator; Bertrand Poincaré Formula; Properties function
1. Introduction
2 z 1 z 1 z mn 1 mn z mn m,n
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.
is called the Weierstrass -function, where mn 2m1 2n2
denotes the sum of all m, n 0, 1, 2, , except for m,nm n 0 . Definition 2 Suppose that L0 is a smooth closed curve in the counterclockwise direction, lying entirely in the fundamental period parallelogram P, with z0 0 and the origin lying in the domain S0 enclosed by L0 . The following operator K a t 0 t0
is called the singular integral operator with -function kernel on L0 , where t H L0 is the unknown function, and
2. Preliminaries
K t0 , t H L0 L0 , a t H L0
Definition 1 Suppose that 1 , 2 are complex constants with Im 1 2 0 , and P denotes the fundamental period parallelogram with vertices 1 2 . Then the K a t 0 t0
b t0 πi
(1) 1 t t0 t0 z0 dt , t0 L0 t K t t ( , ) 0 πi L0
are the given functions. Letting b t K t , t , then (1) becomes
L t t t0 (t0 z0 ) dt 0
1 L K t0 , t K t0 , t0 t t0 t0 z0 t dt πi 0 Copyright © 2013 SciRes.
(2)
AM
L. X. CAO
1232
Since t is uniformly convergent in any closed bounded region lying entirely in P,
obtain K t0 , t K t0 , t0 t t0 t0 z0 N t t0
t t 0 t 0 z 0 1 t t0 M
0 1 , where N is some positive finite constant. Write
for any t0 , t L0 , where M is some positive finite constant. By noting that K t0 , t H 0 1 , we K 0 a t0 t0 k t0 , t
b t0
L t t t0 t0 z0 dt ,
πi
o
1 K t0 , t K t0 , t0 t t0 t0 z0 πi
defined as
D t 1 arg , where S t L 2π 0
k L k t0 , t t dt
S t a t b t , D t a t b t
0
then (1) can be rewritten in the form
K0 k ,
and for definiteness we assume that a 2 t b 2 t 0 , namely we assume that K is an operator of normal type. Now the associated operator of (1) takes the form
(3)
where k is a Fredholm operator and K 0 is called the characteristic operator of K . Now the index of K is
1 K t , t0 t t t0 t0 z0 dt , t0 L, πi L0
K a t0 t0
(4)
or K a t0 t0
1 b t t t t0 t0 z0 dt L k t , t0 t dt , t0 L, 0 πi L0
(4)′
and so that the associated operator of K 0 becomes K 0 a t0 t0
1 b t t t t0 t0 z0 dt , t0 L. πi L0
In addition, if we write 1 k1 t0 L k t , t0 b t b t0 t t0 t0 z0 t dt , t0 L, 0 πi
then (4) can be rewritten as K a t0 t0
b t0 πi
L t t t0 t0 z0 dt k1 t , t0 L,
(5)
0
where b t b t0 t t0 t0 z0 D t t0
So k1 is a Fredholm operator, and then the characK 0 a t0 t0
b t0 πi
( 0 1 , D is some finite constant).
teristic operator of K operator becomes
L t t t0 t0 z0 dt
Therefore, we concluded that K 0 K 0 usually can not be established, that is K 0 K 0 . For convenience, we write z , z z0
where the fixed nonzero point z0 and the origin lie in S0 . It is not difficult to get the following results. Copyright © 2013 SciRes.
(6)
0
Lemma 1 Suppose that f t , H L0 L0 , and with the same L0 as mentioned before, then
a)
L dt L f t, , t d L d L f t, , t dt, 0
0
0
L , t dt L f t , d L 0
0
0
0
d L f t , , t dt 0
AM
L. X. CAO
b) (Poincare-Bertrand formula)
3. Some Properties of Operator K 1) If H , then K H . Proof Through calculation and estimation, we have
2 L t , t0 dt L f t , , t d π f t0 , t0 0
1233
0
L K t1 , t t t1 dt L K t2 , t t t2 dt 0
0
for any t1 , t2 L0 , where M and N are all finite
M t1 t2
N t1 t2
(7)
constant. While for any t1 , t2 L0 , we have
t1 t2 t1 t2 t t 1 2 2 Q t1 t2 t1t2 mn m , n t1 mn t2 mn m,n
t1 t2
where Q is some finite constant. Substituting (8) into (7), we obtain
(8)
L K t0 , t t t0 dt H
(9)
0
Similarly we know that
L K t0 , t t t t0 dt , L K t0 , t t t0 z0 dt , a t0 t0 H 0
0
Consequently, we have K H . 2) If K1 , K 2 are singular integral operator, then K j a j t0 t0
K1 K 2 is also a singular integral operator. That is, if
1 t K j t0 , t t t0 t0 z0 dt , j 1, 2 πi L0
then K1 K 2 a1 t0 a2 t0 b1 t0 b2 t0 t0 a1 t0 K 2 t0 , t a2 t K1 t0 , t t t t0 t0 z0 dt L 0 πi
1
πi
2
L L 0
0
K1 K 2 a1 t0 a2 t0 t0
remainder in that is a Fredholm operator. Proof By definition, we deduce that
1 a1 t0 K 2 t0 , t t t t0 t0 z0 dt πi L0
1 a2 t K1 t0 , t t t t0 t0 z0 dt C t0 πi L0
where C t0
(10)
K1 t0 , t1 K 2 t1 , t t1 t0 t0 z0 t t1 t1 z0 dt1 t dt
where the sum of the former two terms in the right hand of Equation (10) are the characteristic operator, and the
,
1
πi
2
L
0
K1 t0 , t t t0 t0 z0
By virtue of Lemma 1 (b), C t0 can be rewritten in C t0 b1 t0 b2 t0 t0
1
πi
2
L L
Consequently, (10) is established. Copyright © 2013 SciRes.
0
0
L0
K 2 t , t1 t1 t t z0 t1 dt1 dt
the form
K1 t0 , t1 K 2 t1 , t t1 t0 t0 z 0 t t1 t1 z0 dt1 t dt.
Now we write AM
L. X. CAO
1234
L
0
where 1 L
0
K1 t0 , t1 K 2 t1 , t t1 t0 t t1 dt1 1 2 3 4
K1 t0 , t1 K 2 t1 , t
t1 t0 t t1
dt1 ,
t t1
0
t1 t0
0
K1 t0 , t1 K 2 t1 , t
3 L
K1 t0 , t1 K 2 t1 , t
2 L
t t 1 1 2 1 dt1 , mn mn 1 mn
t t m, n
t z 1 1 1 2 0 dt1 , mn 1 t0 mn mn
t m,n
2 2 1 t1 t0 mn 1 mn t1 t0 mn dt1 . 4 L K1 t0 , t1 K 2 t1 , t 1 t t1 mn 1 mn t t1 mn 0 m,n
m, n
By [1], we know that 1 is a Fredholm integral. For 4 , we know from K1 t0 , t1 , K 2 t1 , t ) H L0 L0
that that
4
L
0
is continuous about the variable t L0 , and so 4 t dt is also a Fredholm integral. By noth-
ing that 2 and 3 have the same form, we only need to discuss either one of them. Here we consider the integral 2 . Write 2 h z ' 1 t z mn 1 mn t z mn m,n
then h t1 H L0 is analytic in P and so that h t1 H L0 . Consequently, we read from K1 t0 , t1 , K 2 t1 , t H L0 L0
that 2 H L0 and so that 2 is continuous on L0 , therefore L 2 t dt is also a Fredholm integral. 0 So far, we conclude that K1 K 2 is a singular integral operator. 3) Let K 3 K1 K 2 , where j denotes the indices of K j j 1, 2,3 , then 3 1 2 . Proof From 2), we know
a3 a1a2 b1b2 , b3 a1b2 a2 b1
and S3 S1 S2 , D3 D1 D2
so 3 1 2 . In addition, we can see from a3 a1a2 b1b2 and b3 a1b2 a2 b1 ,
that when K1 , K 2 are normal, K3 is also normal. 4) K1 K 2 K 2 K1 K 2 K 3 . 5) If K is a singular integral operator, and k is a Fredholm integral operator of the first kind, then kK and Kk are also Fredholm integral operators of the first kind. 6) If the indies of K and K are and respectively , then . 7) K1 K 2 K 2 K1 . Through careful calculation, we may obtain 4) - 7). 8) Generally speaking,
L K dt L K dt 0
0
can not be established for , H . Proof By definition and calculation, we have
L K dt L t a t t πi L K t , t1 t1 t1 t t z0 dt1 dt 1
0
0
0
1 L a t t t dt L t L K t , t1 t1 t1 t t z0 dt1dt 0 0 πi 0
.
(11)
Whereas 1
L K dt L a t t t dt πi L t L K t1 , t t1 t1 t t z0 dt1dt . 0
0
0
0
(12)
Let W L t L K t1 , t t1 t1 t t z0 dt1dt 0
then by Lemma 1(a), we have W L
0
L0
0
K t1 , t t t1 t1 t t z0 dt dt1
L t L K t , t1 t1 t1 t t1 z0 dt1 dt 0 0 Copyright © 2013 SciRes.
,
(13)
AM
L. X. CAO
1235
Substituting (13) into (12), we see that 1
L0
K dt a t t tdt i t K t , t1 ) t1 t1 t t1 z0 dt1 dt .
L0
Therefore,
L0
L K dt L K dt 0
0
L0
cannot be established. tion, Vol. 219, No. 4, 2012, pp. 1391-1410. doi:10.1016/j.amc.2012.08.022
REFERENCES [1]
N. I. Muskhelishvili, “Singular Integral Equations,” World Scientific, Singapore City, 1993.
[2]
J. K. Lu, “Boundary Value Problems for Analytic Functions,” World Scientific, Singapore City, 1993.
[3]
F. D. Gakhov, “Boundary Value Problems,” Fizmatgiz, Moscow, 1977.
[4]
M. C. De Bonis and C. Laurita, “Numerical Solution of Systems of Cauchy Singular Integral Equations with Constant Coefficients,” Applied Mathematics and Computa-
Copyright © 2013 SciRes.
(14)
[5]
T. Diogo and M. Rebelo, “Numerical Methods for Nonlinear Singular Volterra Integral Equations,” AIP Conference Proceedings, Kos, 19-25 September 2012, pp. 226-229.
[6]
C. Laurita, “A Quadrature Method for Cauchy Singular Integral Equations with Index −1,” IMA Journal of Numerical Analysis, Vol. 32, No. 3, 2012, pp. 1071-1095. doi:10.1093/imanum/drr032
AM
