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