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# Riemann Boundary Value Problem for Triharmonic Equation in Higher Space

**Academic Editor:**Argiris I. Delis

#### Abstract

We mainly deal with the boundary value problem for triharmonic function with value in a universal Clifford algebra: , , , , , , , where is a Lyapunov surface in , is the Dirac operator, and are unknown functions with values in a universal Clifford algebra Under some hypotheses, it is proved that the boundary value problem has a unique solution.

#### 1. Introduction and Preliminaries

The theory of Riemann boundary value problems in complex plane has been systematically developed in [1, 2]. It is an interesting topic to generalize the classical Riemann boundary value problems theory to Clifford analysis. In [3–6], and so forth, many interesting results about boundary value problem and Riemann Hilbert problems for monogenic functions in Clifford analysis are presented. In [7], Green’s function for the Dirichlet problem for polyharmonic equations was studied. The aim of this paper is to study the Riemann boundary value problem for triharmonic functions. At first, based on the higher order Cauchy integral representation formulas in [8, 9] and the Plemelj formula, we give some properties of triharmonic functions in Clifford analysis, for example, the mean value theorem, the Painlevé theorem, and so forth. Furthermore, on the basis of the above results, we consider the following Riemann boundary value problems: where .

In (1) and (2), and are invertible constants; we denote the inverse elements as and , , , , , . The explicit solutions for (1) are given and the boundary value problem (2) is shown to have a unique solution under some hypotheses.

Let () be an -dimensional real linear space with basis , and the universal Clifford algebra over . For more information on (), we refer to [10–12].

Throughout this paper, suppose is an open, bounded nonempty subset of with a Lyapunov boundary , denoting , . In this paper, for simplicity, we will only consider the case of . The operator is given as

Let be a function with value in , defined in , and the operator acts on the function from the left and from the right, which is being governed by the following rule:

*Definition 1. *A compact surface is called Lyapunov surface with Hölder exponent , if the following conditions are satisfied. (i)At each point there is a tangential space.(ii)There exists a number , such that for any point the set (Lyapunov ball) is connected and parallel lines to the outer normal intersect at not more than one point.(iii)The normal is Hölder continuous on ; that is, there are constants and such that for

Let be an open nonempty subset of with a Lyapunov boundary, , where are real functions; is called a Hölder continuous function on if the following condition is satisfied:
where for any , , , is a positive constant independent of .

Let denote the set of Hölder continuous functions with values in on (the Hölder exponent is , ). We denote the norm in as where

Lemma 2. *The Hölder space is a Banach space with norm (7).*

Denote the fundamental solutions of () by where and denotes the area of the unit sphere in (, , ).

We will introduce the following operators: where .

Lemma 3 (see [5]). *The integral operator in (11) is a bounded linear operator mapping the function space into itself; that is, there exists a positive constant such that, for all ,
*

#### 2. Some Properties for Triharmonic Functions

Theorem 4 (Gauss-mean value formula for triharmonic functions, see [6, 13]). *Suppose in ; then, for any ,
**
or
**
where denotes the area of the unit sphere in .*

Corollary 5. *Suppose in and (); then and in .*

Corollary 6. *Suppose in and is bounded in ; then .*

Theorem 7. *Suppose in , and for , , , and, moreover, , , where . Then in .*

Theorem 8. *Let , in , , (), and (); then for **
where is as in (9) and is a constant.*

*Proof. *For , denote , (). For , denoting
then in . By using Plemelj formula, combining with weak singularity of (), we obtain the following:
where (). By using Theorem 7, we have in . It is clear that we have (). In view of Corollary 6, then the results follow.

Corollary 9. *Let , in , , (), and (), then
*

*Remark 10. *When the condition () in Corollary 9 is replaced by , then the results in Corollary 9 are still valid.

#### 3. Riemann Boundary Value Problem for Triharmonic Functions

In this section, we will consider the Riemann boundary value problem (1); the explicit expression of the solution is given.

Theorem 11. *The Riemann boundary value problem (1) is solvable and the solution can be written as
**
where
*

#### 4. Existence of Solutions for Riemann Boundary Value Problem for Triharmonic Functions

Theorem 12. *Suppose and satisfies the following condition:
**
where is the positive constant mentioned in Lemma 3. Then (2) admits a unique solution.*

*Proof. *Denoting then , . Moreover, by . we get the following:
With being as in (20), it is easy to check that
Denoting , , and using , , we conclude that
where , being as in (26). By Corollary 9, it is clear that , and we then get the following representation formula:
Analogously, we find with from (21) that , . Denote
Using the condition , , we obtain that
where , being as in (27). By Corollary 9, it is clear that , and we then get the following representation formula:

Here being as in (22), then we have , . We denote
and use the condition , . We conclude that
where , being as in (28). Corollary 9 ensures that ; then
Use the same way again, being as in (23), that , . Denoting
and using , , we get
where , being taken from (29). By Corollary 9, it is clear that , and we then get the following representation formula:

Finally, we use as defined in (24) and get , . Define
Working with the condition we arrive at
where , being taken from (30). It is clear that . We obtain that

We only need to consider the existence of solutions to (47). The solution to this problem may be written in the form
where is a Hölder continuous function to be determined on . Then, by using Plemelj formula, (47) can be reduced to an equivalent singular integral equation for ,

Letting denote the integral operator defined by the right hand side of (49), we get
For any , we have
Under the condition (31), the integral operator is a contraction operator mapping the Banach space into itself, which has a unique fixed point for the operator . Thus, there exists a unique solution to (47). The proof is done.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This paper is supported by National Natural Science Foundation of China (11271175), the AMEP, and DYSP of Linyi University.

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#### Copyright

Copyright © 2014 Longfei Gu. 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.