Let be an -connected bounded domain in with the boundary

. Without loss of generality, we assume that is a circular domain

in, bounded by the -circles and. In this article, the notations are the same as in references [1] -[12] . If the first order elliptic system with unknown real functions

satisfies certain conditions, then (1.1) can be transformed into the complex form

where (see Section 4, Chapter 2 in [5] ). Its vector form is as follows:

where is the transposed matrix of. We discuss the first order complex system (1.3) in the form

in which with, with with

We assume (1.4) satisfies the following conditions:

Condition C 1) are continuous in for almost every point

2) The above functions are measurable in for all systems of continuous functions in and any systems of measurable functions in and satisfy

where is as stated in (1.8) below, and are non-negative constants.

3) The complex system (1.4) satisfies the following ellipticity condition

where are non-negative constants.

For convenience, and are used to indicate and respectively, and we define the following:

in which and are stated as in (1.12), (2.1) below, and

and are non-negative constants.

The so-called Riemann-Hilbert boundary value problem for the complex system (1.4) may be formulated as follows.

Problem A Find a system of continuous solutions in of (1.4), which satisfies the boundary condition

in which with for and

are the first kind of discontinuous points of on.

Denote by and the left limit and right limit of as on, and

where when, and when. There is no harm in assuming that the partial indexes of on are not integers, and the partial indexes of on are integers. Set

and we call the index of Problem A.

For problem A, we will assume satisfy the conditions

in which is an open arc from the point to on are non-negative constants,. Moreover, we require that the solution possess the property

in, where in and are small positive constants.

In general, Problem A may not be solvable. Hence we propose a modified problem as follows.

Problem B Find a system of continuous solutions of the complex equation (1.4) in, which satisfies the modified boundary condition

Here

in which are unknown real

constants to be determined appropriately, and, if is an odd integer. More description on and are given below. We begin with the following function

where denotes the partial index on, are fixed points,

which are not the discontinuous points from. Note that the positive direction applies to the boundary circles. Similarly to (1.7)-(1.12), Chapter V, [2] , we see that

Clearly, with certain modification on the symbols on some arcs on, on is seen

to be continuous. In this case, its index

are integers. And we have the following:

in which are solutions of the modified Dirichlet problems with the above boundary conditions for analytic functions, are real constants, and.

In addition, we may assume that the solution satisfies the following point conditions

where are distinct points, and are all real constants satisfying the conditions

for a positive constant. Problem B with in, on and is called Problem.

If then Problem B for (1.4) is the modified Dirichlet boundary value problem for (1.4). It is easy to see that the solutions of (1.4) include the generalized hyperanalytic functions as special cases. In fact, if (1.4) is linear, and and then the solutions of (1.4) are called generalized hyperanalytic functions.

We begin with the following estimates of the solution for problem B.

Theorem 2.1 Suppose that the complex system (1.4) satisfies Condition C and the constants in

(1.6), (1.7), (1.11) are small enough. Then any solution of Problem B for (1.4) satisfies the estimate

where with in,

with are non-negative constants.

Proof There is no harm in assuming that Let It can be seen that is a solution of the following boundary value problem

in which

Following the proof of the Theorem 2.1 of Chapter VI in [1] , we can derive the estimate

From the above estimate, it immediately follows that the estimate (2.1) is true.

In addition, we assume that (1.4) satisfies the following condition: For any continuous vectors and any measurable vector

where satisfy the condition

in which are non-negative constants.

Now, we prove that there exists a unique solution of the modified Riemann-Hilbert problem (Problem B) for analytic vectors by the parameter extensional method.

Theorem 2.2 Let in (1.11) be a sufficiently small positive constant. Then Problem B for analytic vectors has a solution.

Proof We consider the modified Riemann-Hilbert problem (Problem) for analytic vectors with the boundary conditions

where

in which is a real parameter, and is any vector of real functions, and is any

vector of constants. When, it is clear that Problem for analytic vectors has a unique solution (see [1] ). If Problem with for analytic vectors is solvable, we shall prove that there exists a positive number independent of, such that Problem for every has a unique solution. In fact, the boundary conditions (2.9), (2.10) can be rewritten in the form

Substituting the zero vector into the position of on the right hand side of (2.11) and (2.12), by the hypothesis, the boundary value problem (2.11), (2.12) for analytic vectors has a unique

solution and Using the successive iteration,

we can find a sequence of analytic vectors, which satisfies the boundary conditions

From (2.13) and (2.14), we have

In accordance with Theorem 2.1, we can conclude

where with, and Choosing a positive constant, such that it is not difficult to see that

and

for where is a positive integer. This shows that

Hence, there exists an analytic vector such that

Thus is a solution of Problem with. From this we can derive that Problem with i.e. Problem B for analytic vectors is solvable.

Next we prove the solvability of Problem B for the system (1.4).

Theorem 2.3 Let the nonlinear elliptic system (1.4) satisfy Condition C, and in (1.6), (1.7), (1.11) be sufficiently small positive constants. Then Problem B for the complex system (1.4) is solvable.

Proof We consider the nonlinear elliptic complex system with the parameter:

where is any measurable vector in and Applying Theorem 2.2, we see that Problem B for (2.19) with is

solvable, and the solution can be expressed as

where is an analytic vector satisfying the boundary conditions

Suppose that when, Problem B for the system (2.19) has a unique solution. Then we shall prove that there exists a neighborhood of so that for every and any function Problem B for (2.19) is solvable. In fact, the complex system (2.19) can be written in the form

Suppose that Problem B for (2.13) with is solvable, by using the similar method as in the proof of Theorem 2.2, we can find a positive constant, so that for every, there exists a sequence of solutions satisfying

The difference of the above equations for and is as follows:

From Condition C, we can derive that

and

Moreover, satisfies the homogeneous boundary conditions

Similarly to Theorem 3.3, Chapter I, [1] , we have

where are positive constants. Provided is small enough, so that we can obtain

for every Thus

for where is a positive integer. This shows that as Thus there exists a system of continuous functions in, such that

By Condition C, it follows that is a solution of Problem B for the system (2.23), i.e. (2.19) for. It is easy to see that the positive constant is independent of. Hence Problem B for the system (2.19) with is solvable. Correspondingly we can derive that when, Problem B for (2.19) is solvable. Especially Problem B for (2.19) with and, namely Problem B for the system (1.4) has a solution.

In this section, we shall introduce an error estimate of the above approximate solutions.

Theorem 3.1 Under the same conditions as in Theorem 2.3, let be a solution of Problem B for the complex system (1.4) satisfying Condition C in, and be its approximation as stated in the proof of Theorem 2.3 with Then we have the following error estimate

where with as in (2.28), and as in (1.6),(1.7), (1.11) and (1.16).

Proof From (1.4) and (2.24) with, we have

It is clear that satisfies the homogeneous boundary conditions

Noting that satisfy, and

and then is a solution of Problem for the complex equation

hence we have

in which

where the non-negative constants are as stated in (2.28), (1.5), (1.11) and (1.12). Moreover according to the proof of Theorem 2.3, we can derive

From (3.6) and (3.7), it follows that

where and is the solution of Problem B for (2.24) with and Finally, we obtain

This shows that (3.1) holds. If the positive constant is small enough, so that when is sufficiently large and is close to 1, then the right hand side becomes very small.

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