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

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.


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][2][3][4][5][6][7][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][10][11][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.

A Riemann Boundary Value Problem of Non-Normal Type on the Infinite Straight Line
Let X be the real axis oriented in the positive direction.
And let Z  , Z  denote the upper half-plane and the lower half-plane cut by X .Our objective is to find a sectioally holomorphic function satisfying the following boundary condition where the two given functions X with existing and (clearly exists), and being on X and , Without loss of generality, we can consider problem (1) in class 0 , that is, the two limits Clearly, here we have

Homogeneous Problem
The homogeneous problem of (1) is as follows (2) then and

Since
, by taking logarithm of for some branch we obtain a single-valued function with , hence . And by simple calculation we see that then then we get . Thus is analytic on the whole complex plane and has at most . From [5], we know that must be an arbitrary polynomial of degree . Therefore, the homogeneous problem (2) has general solution in class as follows . where

Now we get the general solution in class
for the homogeneous problem (2) as follows , .
Thus we get the following results.Theorem 3.1.For the homogeneous problem (2) in class , the following two cases arise.
 , 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.

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

 
X z is given by ( 6) but with By this, problem (1) can be rewritten as We note that Plemelj formula can not be used directly here, because that when 0 is not a finite constant, and so (unless 0   ).For a unified treatment, regardless of the value of  , we always let to the two sides of (9) gives and so that then we get and with . Similar to the reasoning for (7) for problem (2), we know that if problem (1) has solution in class 0 , 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.
F z has a pole of order  at .To eliminate the singularity, we multiply to F z and get a polynomial of degree : , is actually the general solution for problem (1) in class 0 , where .For convenience, we deform the function given by (12) into are bounded on X , and the fact that in the numerator of the above formula should be  , we suppose that

 
x   is bounded on X ) only when In a similar way, we know that are satisfied.Hence, we get the results that are bounded on X 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 P coefficients by using H Firstly, we make the polyn of degree ermite interpolation polynomial.omial The polynomial are continuous on X , and through simple verifi that the condition in problem (1), and that the order of we aim to m ow ake and is justly the solution eo   0 z  for ho longs mogeneous problem f (2), and also a particular solution for nonhomogeneous problem (1) in 0 R .Combining with the general solution (8) of homogen us problem (2), we known that when here when , ; e e e , Here we write Since the order of the denominator in is satisfied, (16) actually the general solution for nonhomogenous problem (1), now the homogeneous problem (2) only has Case 2.
. 0 When is analytic on the whole complex plane and has  order at  , that is, , and now the ho ogeneous pr olution.Therefore the general solution for nonhomogenous problem (1) in 0 R is given by m oblem (2) only has zero-s e , ( ) If we make the Hermite interpolation polynomial of degree