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We consider a Hilbert boundary value problem with an unknown parametric function on arbitrary infinite straight line passing through the origin. We propose to transform the Hilbert boundary value problem to Riemann boundary value problem, and address it by defining symmetric extension for holomorphic functions about an arbitrary straight line passing through the origin. Finally, we develop the general solution and the solvable conditions for the Hilbert boundary value problem.

Various kinds of boundary value problems (BVPs) for analytic functions or polyanalytic functions have been widely investigated [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 holomorphic functions or bianalytic functions have been investigated [9-12].

In this paper, we consider a kind of Hilbert BVP with an unknown parametric function. We first define the symmetric extension of holomorphic function about an infinite straight line passing through the origin, and discuss its several important properties. And after, we propose a Hilbert BVP with an unknown parametric function on arbitrary half-plane with its boundary passing through the origin. Then, we transform the Hilbert BVP into a Riemann BVP on the infinite straight line using the defined symmetric extension. Finally, we discuss the solvable conditions and the solution for the Hilbert BVP.

Let be an infinite straight line with an inclination in the complex plane, passing through the origin and being oriented in upward direction. Let and denote the upper half-plane and the lower halfplane cut by.

Our objective is to find a pair of functions , where is holomorphic in the domain and continuously extendable to its boundary, and is real-valued and Holder continuous on, satisfying the following boundary conditions

where

and

are given functions.

An important step in solving problem (1) is to define a symmetric extension of holomorphic functions about the infinite straight line with an inclination.

For a holomorphic function in the simplyconnected domain, we define the symmetric extension of about as follows:

where is the symmetric point of about. For simplicity, we express as. From definition (2), we may establish that 1);

2) If is defined in, then is also defined in;

3) If is holomorphic in, then is holomorphic in because of

;

4) If a holomorphic in can be continuously extended to, then in can be continuously extended to, and their boundary value on satisfies the following equality

; (3)

5) If is holomorphic in and continuous on, then

is a sectionally holomorphic function that jumps on with finite, and possesses the following properties:

6) Let, where and are holomorphic in (or). It is not necessarily true that.

Problem (1) is normal only if on.

In this section, we develop a general method to solve boundary value problem (1) or similar problems. Let

Multiplying the first and the second equation in (1) by and respectively, we obtain the Riemann boundary problem

or

By extending to about the straight line, we obtain a sectionally holomorphic function as (4) with jump, satisfying the boundary conditions

.

Thus (9) can be rewritten in the form

Due to, (10) can be written as R problem

, (10)’

where

and, on.

If is a solution of problem (1), then extended from by (4) must be a solution of (10) or (10)’ in (namely) and satisfies the boundary condition (5). On the other hand, if the solution of R problem (10)’ in satisfies the boundary condition (5), then is really a solution of problem (1). Consequently, problem (1) is equivalent to R problem (10)’ in together with the additive condition (5).

Assume that is a solution of (10) in, by making conjugate for (10) we obtain

.

We read from relation (7) that is also a solution of (10)’ in, so that

is a solution of (10)’ in class and satisfies the additive condition (5). So that, whenever we find out the solution of problem (10)’ in class, and write out, then

is actually the solution of problem (1).

Let

.

By we know that is even.

Here, we only consider the problem (1) in the normal case. The nonnormal case can be solved similarly.

The homogeneous problem of (1) is as follows

By canceling the unknown function, problem (12) becomes

which corresponds to the homogeneous problem of R problem (10)’

Setting, we have

.

If we let, then we know on with

, and. By letting

we can rewrite (14) as follows

Let us introduce the function

.

Since, we have

where

is real-valued on and. Now the canonical function of R problem (15) or (14) can be taken as

where is an unknown complex constant. We can see from (17) that, thus R problem (15) can be transferred to the following problem

So is holomorphic on the whole complex plane and has order at. From [

where is an arbitrary polynomial of degree with if.

According to (16), we know

and hence

From (17) and (20) it can be seen that

which implies that. By taking

we obtain

and

Consequently, we see that if and only if

Then when condition (24) is satisfied, the solution of H problem (13) is given by (19).

Now putting the solution of H problem (13) given by (19) into the first equation (or the second equation) in (12), we get

Thus we get the following results.

Theorem 5.1. For the homogeneous problem (12), the following two cases arise.

1) When, its general solution is, where and are given by (19) and (25) respectively, in which condition (24) is satisfied for, and is given by (22) (a real constant factor is permitted for).

2) When, it only has zero-solution

.

In order to solve the n nonhomogeneous problem (1), we only need to find out a particular solution for problem (1).

According to [

Therefore is actually the particular solution of problem (1), where

And from (20) we obtain

It follows from (3) that and from (20) and (21) that

.

While due to and (28) we have, so we obtain

Therefore, we obtain

and

When, has singularity of order at. Now we aim to cancel the singularity of at. From [

and its unique solution takes the form

For the case, since the solution for (10)’ in is unique and must be a solution of (10)’ in, we conclude that, thus (33) is actually the unique solution of nonhomogeneous problem (8).

Combining the particular solution of nonhomogeneous problem (8) and the general solution of homogeneous problem (14), we known that when, the general solution of R problem (8) is

where satisfies condition (24) and is given by (31); when, R problem (8) is solvable if and only if (32) is satisfied and the unique solution is given by (33).

Putting the solution into the first equation in (1), we obtain

Therefore, we derive the following results.

Theorem 5.2. If, the nonhomogeneous problem (1) is always solvable and its general solution is, where is given by (34) with satisfying condition (24) and being given by (22) (a real constant factor is permitted for), while is given by (35). If, under the necessary and sufficient condition (32), the nonhomogeneous problem (1) has unique solution, where and are given by (33) and (35) respectively.