LI Kaiya，LIU Hua，WEI Xin,QU Feifei

(1.School of Science，Tianjin University of Technology and Education，Tianjin 300222;2.School of Science，Xi′an Shiyou University，Xi′an Shaanxi 710065)

Abstract The direct method of solution for singular integral equations with Hilbert kernel is discussed in this paper.Following the route of the direct method of solution for singular integral equations with Cauchy kernel，the generalized residue theorem and Plemelj formula in periodic form are stated.Then the singular integral equation with Hilbert kernel is directly transformed into an algebraic equation without the numerical discretization of the singular integral.At last the equivalence between algebraic equation and original equation is proved.

Key words Singular integral equation with Hilbert kernel;Generalized residue theorem;Plemelj formula;Direct method

As an important tool of modern applied mathematics，singular integral equation appears widely in many theoretical research and engineering practices.

The singular integral equation with Cauchy kernel is represented as

(1)

In the early applications of singular integral equations，especially in fracture mechanics，it is the conventional practice for the direct discretization of the equation.However，it is not easy to discretize the integral with singular kernel because of the low efficiency for the most of traditional calculation of the algebraic equation obtained by discretization.Therefore，after the emergence of the finite element method，this method (direct method of numerical solution) is not commonly used.During the 1980s and 1990s，a large number of new methods have been developed to deal with the numerical solutions of singular integral equations.These algorithms，which mainly based on collocation method，often have some further restrictions ona(t) andb(t).So far，the numerical solutions of singular integral equations have still not been fully adopted in practice.

There is an effective method to deal with singular integral equations in some special cases，which is called the direct method.In 1965，A.S.Peters discussed that when the coefficienta(t) and the kernel densityK(t,τ) are holomorphic functions，the singular integral equation does not need to be transformed into Fredholm equation also not into Riemann boundary value problems，while the close form of the solution is obtained[1].Later，K.M.Case (1966)，S.G.Samko (1969)，and F D Gakhov (1977)，continued to discuss this question.But their solvable conditions are too complicated to check[2-4]. In 1975，Lu Jianke simplified Peters' method by removing some unnecessary restrictions.In more general cases，he obtained the solution and solvable conditions in an obvious close form，so it is convenient for applications[5-7].

In this paper，we introduce the direct method to the singular integral equation with Hilbert kernel(SIEH)，which is the essential tool in the theory of the elasticity in the periodic complex plane[8].We give the general model of SIEH as follows[9]

(2)

Now，we discuss the principal part of (2)，i.e.,its characteristic equation

(3)

Recall that the fundamental tool of the direct method for singular integral equations with Cauchy kernel is the generalized residue theorem.Accordingly，the direct method of solution for (3) also depends on its periodic form.

Different from the Cauchy integral on a closed curve，to deal with the generalized residue theorem of the integral

we need to consider the residue atz=±i.LetH2πbe the H?lder continuous function space with period of2π.

Figure1periodicstripFigure2basicdomainD

Without confusion，from now on we denote byDthe fundament stripD1.LetLbe an arbitrary arc onD(Figure 2).

Lemma1 (The generalized Plemelj formula)

Letg(t)∈Hwith period of 2πand

(4)

then

(5)

1 Periodic Form of the Generalized Residue Theorem

Theorem1 Let periodic functionh(z) be bounded at infinity.h(z) has isolated singularitiesz1,z2,…,zMinD.Ifh(z) has polest1,t2,…,tNonL1{0} andx1,x2,…,xKon [0,2π) of ordern1,n2,…,nN,k1,k2,…，kKrespectively，besides

(6)

(i).

(7)

ProofLetYis big enough so thath(z) has no pole outside of the regionDas shown in Figure 2.

(8)

Here，the singular point of periodic functionh(z) may be onL1，[0,2π]，L3orD+.Assume thath(z) has isolated singularitiesz1,z2,…,zMinD+and polest1,t2,…,tNonL1{0} of ordern1,n2,…,nNrespectively，x1,x2,…,xKon [0,2π) of orderk1,k2, …，kKrespectively.Therefore，according to (8)：

(i) Whenx1=0，it follows that:

(9)

whereh(tj)=h(tj+2π) andh(0)=h(2π)，becauseh(z) is a function with a period of 2π.

(ii) Whenx1∈(0,2π)，because of

Resh(0)=Resh(2π)=0,

(9) is still valid.

Secondly,we consider the left side of (9).By Figure 2，we have

(10)

becauseh(z) may have poles on the boundary ofD，this integral perhaps is a higher order singular integral[7].Sinceh(z) has a period of2π，we obtain that

Therefore (10) can be written as

(11)

Sinceh(z) is a periodic function and bounded at infinity,its limit exists at the infinite point. Therefore，whenY→，we have

thus (11) can be written as

(12)

Finally,combining (9) and (12),we obtain that

2 Presentation of Singular Integral Equation with Hilbert Kernel

We discuss the characteristic equation of (2):

(13)

In this article we restrict ourselves to the case of normal type:

A2(t)-B2(t)≠0，t∈R.

3 Main Result

We deal with the problem (13) by the similar way to the singular integral equation with Cauchy kernel[5-7].LetA(z),B(z) andf(z) be as above.

The key to solve the problem is the distribution of the zeros ofA(z)±B(z) inD+.We can assume that

(i)A(z)+B(z) has zerosα1,α2,…,αmof ordersλ1,λ2,…,λmrespectively.

(ii)A(z)-B(z) has zerosβ1,β2,…,βnof ordersμ1,μ2,…,μnrespectively.

(αkandβjmay be coincident) inD+,

then

The index of (13) is defined by

Assume prior (13) has a solutionφ(t).

Define by

(14)

then Φ(z) is a sectional holomorphic function with period of2π.By Lemma 1，we have

(15)

which also belongs toH.Substituting (15) into (13)，we obtain that

(16)

Substituting (16) back into (14)，we get

(17)

where

that is a known function holomorphic inD+,and by Theorem 1

=I1+I2+I3.

(We have putz≠βj,let Φ(βj)is the limit value whenz→βj)

Now we have

and

in which we have put

and

At last，since

and the integrand of (14) is bounded near+i，we get

Φ(i)

Therefore

where

TakingI1into (17)，we have

that is

(18)

SubstitutingI2,I3back into (18)，we obtain that

(19)

Evidently，to guarantee that the equation (13) is solvable，it is necessary that the following two conditions must be satisfied:

1°.Whenz=βj(j=1,2,…,n) is substituted into (19) and up to its derivative of orderμj-1，there appears no contradiction between the two sides.

2°.The right side of (19) is indeed a holomorphic function inD+.

LetCjr=Φ(r)(βj) be undetermined constants.By the two requirements，we obtain the linear algebraic equations {Cjr}.The compatibility condition of the linear equations {Cjr} is the necessary condition for (13) to be solved.Let’s prove that it is also a sufficient condition.

Replace Φ(r)(βj) in (19) by the obtained solutionsCjr:

(20)

Taking its boundary value asz→t∈L(by Lemma 1，that the Plemelj formula is valid forF+(t) as well as Φ+(t)∈H)，we obtain that

Substituting it into (16)，we obtain that

(21)

Let

Integrating on both side of (21)，we obtain that

where

and

Again by Theorem 1

Case1WhenC2≠1，we can calculate the value ofλ:

thus (21) can be written by

(22)

Case2WhenC2=1，we needC1=0 and then it has the solution

(23)

whereCis an arbitrary constant.

As above，when the linear equations {Cjr} are compatible，(22) or (23) is the general solution of (13).Hence，we have proved that the compatibility condition of {Cjr} is a necessary and sufficient solvable conditions of (13).

Now we obtain the last result in this article.