** Applied Mathematics ** Vol. 4 No. 8 (2013) , Article ID: 35589 , 5 pages DOI:10.4236/am.2013.48165

Riemann Boundary Value Problem of Non-Normal Type on the Infinite Straight Line

Department of Information and Computing Sciences, Mathematics College, Northeast Petroleum University, Daqing, China

Email: caolixia98237@163.com

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.

Received June 16, 2013; revised July 16, 2013; accepted July 25, 2013

**Keywords:** Non-Normal Type; Riemann Boundary Value Problem; The Infinite Straight Line

**ABSTRACT**

**We consider a Riemann boundary value problem of non-normal type on the infinite straight line. By using the method of complex functions, we investigate the method for solving this Riemann boundary value problem of non-normal type and give the general solutions and the solvable conditions for it.**

1. Introduction

Various kinds of Riemann boundary value problems (BVPs) for analytic functions on closed curves or on open arc, doubly periodic Riemann BVPs, doubly quasi-periodic Riemann BVPs, and BVPs for polyanalytic functions have been widely investigated in [1-8]. The main approach is to use the decomposition of polyanalytic functions and their generalization to transform the boundary value problems to their corresponding boundary value problems for analytic functions. Recently, inverse Riemann BVPs for generalized analytic functions or bianalytic functions have been investigated in [9-12].

In this paper, we consider a kind of Riemann BVP of non-normal type on the infinite straight line and discuss the solvable conditions and the general solution for it.

2. A Riemann Boundary Value Problem of Non-Normal Type on the Infinite Straight Line

Let be the real axis oriented in the positive direction. And let, denote the upper half-plane and the lower half-plane cut by. Our objective is to find a sectioally holomorphic function satisfying the following boundary condition

(1)

where the two given functions and on with existing and (clearly exists), and

, with

being on and. And the integer

is the index of problem (1).

Write

.

Without loss of generality, we can consider problem (1) in class, that is, the two limits and exist as. Clearly, here we have.

3. Homogeneous Problem

The homogeneous problem of (1) is as follows

. (2)

It is found that is required for solving problem (2) in class. Here we suppose that. Let

, (3)

then and with

and

. (4)

Write

. (5)

Since, by taking logarithm of for some branch we obtain a single-valued function with, hence exists with and. And by simple calculation we see that is sectionally holomorphic.

Write

(6)

then. Substituting this into (2) gives

.

If we write

then we get. Thus is analytic on the whole complex plane and has at most order at. From [5], we know that must be an arbitrary polynomial of degree with if. Therefore, the homogeneous problem (2) has general solution in class as follows

(7)

Considering the requirements that and are bounded at

we can let

where is an arbitrary polynomial of degree with if. Now we get the general solution in class for the homogeneous problem (2) as follows

(8)

Thus we get the following results.

Theorem 3.1. For the homogeneous problem (2) in class, the following two cases arise.

1) When, it is always solvable and its general solution is given by (8), where is an arbitrary polynomial of degree.

2) When, it only has zero-solution.

4. Nonhomogeneous Problem

For nonhomogeneous problem (1), the key is to find out the special solution.

Similar to the case in homogeneous problem (2), the canonical function is given by (6) but with

satisfying

.

By this, problem (1) can be rewritten as

(9)

We note that Plemelj formula can not be used directly here, because that when

is not a finite constant, and so (unless

). For a unified treatment, regardless of the value of, we always let

Multiplying to the two sides of (9) gives

.

We know that and so that

.

If we let

, (10)

then we get

and

(11)

with. Similar to the reasoning for (7) for problem (2), we know that if problem (1) has solution in class, then can easily write out the form. But for problem (11), the function

is analytic everywhere except at the possible unique pole, therefore the following two cases arise.

Case 1..

When, has a pole of order at. To eliminate the singularity, we multiply to and get a polynomial of degree:

.

Therefore

(12)

is actually the general solution for problem (1) in class, where. For convenience, we deform the function given by (12) into

(13)

Considering the requirements that are bounded on, and the fact that

with, the smallest degree of in the numerator of the above formula should be, we suppose that

then when, we have

and the smallest degree of in numerator is (i.e. is bounded on) only when

that is,

(14)

In a similar way, we know that is bounded on only when

(15)

are satisfied.

Hence, we get the results that are bounded on only when the conditions (14) and (15) are all satisfied. While it is troublesome to solve the system composed by (14) and (15) for the coefficients of.

Here we aim to determine the coefficients by using Hermite interpolation polynomial.

Firstly, we make the polynomial of degree such that

The polynomial exists uniguely from [5]. Let

(16)

We can see from (16) that are continuous on, and through simple verification that satisfy the condition in problem (1), and that the order of at is.

Now we aim to make belong to by adding restricted conditions.

a) If, that is, , exactly belongs to and is justly the solution for homogeneous problem of (2), and also a particular solution for nonhomogeneous problem (1) in. Combining with the general solution (8) of homogeneous problem (2), we known that when, the general solution of problem (1) in is

(17)

where when.

b) If, that is, , since the homogeneous problem (2) corresponding to (1) only has zero-solution in, the existence of a solution for nonhomogeneous problem (1) in implies the uniqueness. Because has generally singularity of order at, belongs to if and only if conditions on or on are satisfied.

By rewriting as

Here we write

. (18)

Since the order of the denominator in is (is always true), it is enough that is at most a polynomial of degree, that is,

. (19)

Therefore, only when condition (19) is satisfied, (16) is actually the general solution for nonhomogenous problem (1), now the homogeneous problem (2) only has zero-solution.

Case 2..

When, is analytic on the whole complex plane and has order at, that is, , and now the homogeneous problem (2) only has zero-solution. Therefore the general solution for nonhomogenous problem (1) in is given by

(20)

If we make the Hermite interpolation polynomial of degree, the fact that (the order of the denominator) implies that is unbounded on, which contradicts with the hypothesis that is bounded on. So it is infeasible to make the Hermite interpolation polynomial for this.

However, we have the following effective treatment for this.

a) Under this situation, may have the unique pole at. In order to eliminate the pole, we should put the following restrictions for it:

if, we only need

; (21)

if, we only need to put

; (22)

if, apart from (22), the restrictions

or

(23)

are necessary.

b) Considering the boundedness of on, the following restrictions are also necessary:

(24)

(25)

Thus we get the following results.

Theorem 4.1. For the nonhomogeneous problem (1) in class, the following two cases arise.

1). If, problem (1) is always solvable and its general solution is given by (17); and if, if and only if satisfies condition (19), problem (1) has unique solution, given by (16).

2) When, it has unique solution in form (20) when the restrictions (21) or (22) or (22)-(23), and (24)- (25) are satisfied.

Anyway, the degree of freedom of solution for nonhomogeneous problem (1) is.

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