# Riemann Boundary Value Problem for Triharmonic Equation in Higher Space

Authors: Longfei Gu
Journal: The Scientific World Journal
 (2014)

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## Abstract

We mainly deal with the boundary value problem for triharmonic function with value in a universal Clifford algebra:Δ3[u](x)=0, x∈Rn∖∂Ω, u+(x)=u-(x)G(x)+g(x), x∈∂Ω, (Dju)+(x)=(Dju)-(x)Aj+fj(x), x∈∂Ω, u(∞)=0, where (j=1,…,5)  ∂Ω is a Lyapunov surface in Rn, D=∑k=1nek(∂/∂xk) is the Dirac operator, and u(x)=∑AeAuA(x) are unknown functions with values in a universal Clifford algebra Cl(Vn,n). Under some hypotheses, it is proved that the boundary value problem has a unique solution.


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## Body

## 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:
(1)Δ3[u](x)=0,x∈Rn∖∂Ω,u+(x)=u-(x)A+g(x),x∈∂Ω,(Dju)+(x)=(Dju)-(x)Aj+fj(x),x∈∂Ω,|u(∞)|≤C*,(2)Δ3[u](x)=0,x∈Rn∖∂Ω,u+(x)=u-(x)G(x)+g(x),x∈∂Ω,(Dju)+(x)=(Dju)-(x)Aj+fj(x),x∈∂Ω,u(∞)=0,
where (j=1,…,5).In (1) and (2), A and Aj are invertible constants; we denote the inverse elements as A-1 and Aj-1.u(x), (Dju)(x), g(x), fj(x)∈Hβ(∂Ω,Cl(Vn,n)), j=1,…,5, 0<β≤1. The explicit solutions for (1) are given and the boundary value problem (2) is shown to have a unique solution under some hypotheses.LetVn,s (0≤s≤n) be an n-dimensional (n≥1) real linear space with basis {e1,e2,…,en}, and Cl(Vn,s) the universal Clifford algebra over Vn,s. For more information on Cl(Vn,s) (0≤s≤n), we refer to [10–12].Throughout this paper, supposeΩ is an open, bounded nonempty subset of Rn with a Lyapunov boundary ∂Ω, denoting Ω+=Ω, Ω-=Rn∖Ω¯. In this paper, for simplicity, we will only consider the case of s=n. The operator D is given as
(3)D=∑k=1n‍ek∂∂xk:Cr(Ω,Cl(Vn,n))⟶Cr-1(Ω,Cl(Vn,n)).Letu be a function with value in Cl(Vn,n), defined in Ω, and the operator D acts on the function u from the left and from the right, which is being governed by the following rule:
(4)D[u]=∑k=1n‍∑A‍ekeA∂uA∂xk,[u]D=∑k=1n‍∑A‍eAek∂uA∂xk.Definition 1.
A compact surfaceΓ is called Lyapunov surface with Hölder exponent α, if the following conditions are satisfied. (i)
At each pointx∈Γ there is a tangential space.(ii)
There exists a numberr, such that for any point x∈Γ the set Γ∩Br(x) (Lyapunov ball) is connected and parallel lines to the outer normal α(x) intersect at not more than one point.(iii)
The normalα(x) is Hölder continuous on Γ; that is, there are constants C>0 and 0<α≤1 such that for x,y∈Γ(5)|α(x)-α(y)|≤C|x-y|α.
LetΩ be an open nonempty subset of Rn with a Lyapunov boundary, u(x)=∑AeAuA(x), where uA(x) are real functions; u(x) is called a Hölder continuous function on Ω¯ if the following condition is satisfied:
(6)|u(x1)-u(x2)|=[∑A‍|uA(x1)-uA(x2)|2]1/2≤C|x1-x2|α,
where for any x1,x2∈Ω¯, x1≠x2, 0<α≤1, C is a positive constant independent of x1,x2.LetHα(∂Ω,Cl(Vn,n)) denote the set of Hölder continuous functions with values in Cl(Vn,n) on ∂Ω (the Hölder exponent is α, 0<α<1). We denote the norm in Hα(∂Ω,Cl(Vn,n)) as
(7)∥u∥(α,∂Ω)=∥u∥∞+∥u∥α,
where
(8)∥u∥∞∶=sup⁡x∈∂Ω⁡|u(x)|,∥u∥α∶=sup⁡x1,x2∈∂Ωx1≠x2|u(x1)-u(x2)||x1-x2|α.Lemma 2.
The Hölder spaceHα(∂Ω,Cl(Vn,n)) is a Banach space with norm (7).Denote the fundamental solutions ofDj (j=1,2,…,6) by
(9)H1(x)=1ωnxρn(x),H2(x)=12-n1ωn1ρn-2(x),H3(x)=12(2-n)1ωnxρn-2(x),H4(x)=12(2-n)(4-n)1ωn1ρn-4(x),H5(x)=18(2-n)(4-n)1ωnxρn-4(x),H6(x)=18(2-n)(4-n)(6-n)1ωn1ρn-6(x),
where ρ(x)=(∑i=1n‍xi2)1/2 and ωn denotes the area of the unit sphere in Rn (n≥3, n≠4, n≠6).We will introduce the following operators:(10)(F∂Ωu)(x)=∫∂Ω‍2H1(y-x)dσyu(y),x∈Rn∖∂Ω,(11)(S∂Ωu)(x)=∫∂Ω‍2H1(y-x)dσyu(y),x∈∂Ω,
where u∈Hα(∂Ω,Cl(Vn,n)).Lemma 3 (see [5]).
The integral operatorS∂Ω in (11) is a bounded linear operator mapping the function space Hα(∂Ω,Cl(Vn,n)) into itself; that is, there exists a positive constant M such that, for all u∈Hα(∂Ω;Cl(Vn,n)),
(12)∥S∂Ωu∥(α,∂Ω)≤M∥u∥(α,Cl(Vn,n).


## 2. Some Properties for Triharmonic Functions

Theorem 4 (Gauss-mean value formula for triharmonic functions, see [6, 13]).
SupposeΔ3[u]=0 in Rn; then, for any x∈Rn,
(13)u(x)=1ωnRn-1∫∂B(x,R)‍u(y)dS-R22nΔ[u](x)-R48n(n+2)Δ2[u](x),
or
(14)u(x)=nωnRn∫B(x,R)‍u(y)dV-R22(n+2)Δ[u](x)-R48(n+2)(n+4)Δ2[u](x),
where ωn denotes the area of the unit sphere in Rn.Corollary 5.
SupposeΔ3[u]=0 in Rn and |u(x)|=O(|x|) (|x|→∞); then Δ[u]=0 and Δ2[u]=0 in Rn.Corollary 6.
SupposeΔ3[u]=0 in Rn and u(x) is bounded in Rn; then u(x)≡C.Theorem 7.
SupposeΔ3[u]=0 in Rn∖∂Ω, and for x∈∂Ω, u∈C6(Rn∖∂Ω,Cl(Vn,n)), [u]+(x)=[u]-(x)∈C5(∂Ω,Cl(Vn,n)), and, moreover, Dj[u]+(x)=Dj[u]-(x)∈Hαj(∂Ω,Cl(Vn,n)), 0<αj≤1, where j=0,1,…,5. Then Δ3[u]=0 in Rn.Theorem 8.
Letu∈C6(Ω-,Cl(Vn,n))∩C5(Ω-¯,Cl(Vn,n)), Δ3[u]=0 in Ω-, Dj[u](x)∈Hαj(∂Ω,Cl(Vn,n)), 0<αj≤1 (j=0,1,…,5), and |u(x)|=O(1) (|x|→∞); then for x∈Ω-(15)u(x)=∑j=05‍(-1)j+1∫∂ΩHj+1(y-x)dσyDj[u](y)+C,
where Hj(y-x) is as in (9) and C is a constant.Proof.
Fory∈∂Ω, denote Dj[u](y)=-Ψj(y), (j=0,1,…,5). For x∈Rn∖∂Ω, denoting
(16)Θ(x)=∑j=05‍(-1)j∫∂ΩHj+1(y-x)dσyΨj(y),u*(x)={-Θ(x),x∈Ω+u(x)-Θ(x),x∈Ω-,
then Δ3[u*]=0 in Rn∖∂Ω. By using Plemelj formula, combining with weak singularity of Hj(x) (j≥2), we obtain the following:
(17)Dj[u*]+(x)=Dj[u*]-(x)∈Hαj′(∂Ω,Cl(Vn,n)),
where 0<αj′≤1 (j=0,1,…,5). By using Theorem 7, we have Δ3[u*]=0 in Rn. It is clear that we have |u*(x)|=O(1) (|x|→∞). In view of Corollary 6, then the results follow.Corollary 9.
Letu∈C6(Ω-,Cl(Vn,n))∩C5(Ω-¯,Cl(Vn,n)), Δ3[u]=0 in Ω-, Dj[u](x)∈Hαj(∂Ω,Cl(Vn,n)), 0<αj≤1 (j=0,1,…,5), and |u(x)|=O(1) (|x|→∞), then
(18)Dj[u](∞)=0,j=1,…,5,Remark 10.
When the condition|u(x)|=O(1) (|x|→∞) in Corollary 9 is replaced by u(∞)=0, 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
(19)u(x)={∑i=16‍Ψi(x)+C,x∈Ω,∑i=16‍Ψi(x)+CA-1,x∈Ω-,
where
(20)Ψ1(x)={-∫∂Ω‍H6(y-x)dσyf5(y),x∈Ω+-∫∂Ω‍H6(y-x)dσyf5(y)A5-1,x∈Ω-(21)Ψ2(x)={∫∂Ω‍H5(y-x)dσyf~4(y),x∈Ω+∫∂Ω‍H5(y-x)dσyf~4(y)A4-1,x∈Ω-(22)Ψ3(x)={-∫∂Ω‍H4(y-x)dσyf~3(y),x∈Ω+-∫∂Ω‍H4(y-x)dσyf~3(y)A3-1,x∈Ω-(23)Ψ4(x)={∫∂Ω‍H3(y-x)dσyf~2(y),x∈Ω+∫∂Ω‍H3(y-x)dσyf~2(y)A2-1,x∈Ω-(24)Ψ5(x)={-∫∂Ω‍H2(y-x)dσyf~1(y),x∈Ω+-∫∂Ω‍H2(y-x)dσyf~1(y)A1-1,x∈Ω-(25)Ψ6(x)={∫∂Ω‍H1(y-x)dσyg~(y),x∈Ω+∫∂Ω‍H1(y-x)dσyg~(y)A-1,x∈Ω-(26)f~4(x)=f4(x)-∫∂Ω‍H2(y-x)dσyf5(y)(-1+A5-1A4),hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhihx∈∂Ω,(27)f~3(x)=f3(x)+∫∂ΩH3(y-x)dσyf5(y)(-1+A5-1A3)-∫∂ΩH2(y-x)dσyf~4(y)(-1+A4-1A3),mmmmmmmmmmmmmmmmmmmmx∈∂Ω,(28)f~2(x)=f2(x)-∫∂ΩH4(y-x)dσyf5(y)(-1+A5-1A2)+∫∂ΩH3(y-x)dσyf~4(y)(-1+A4-1A2)-∫∂ΩH2(y-x)dσyf~3(y)(-1+A3-1A2),mmmmmmmmmmmmmmimmmmmmx∈∂Ω,(29)f~1(x)=f1(x)+∫∂ΩH5(y-x)dσyf5(y)(-1+A5-1A1)-∫∂ΩH4(y-x)dσyf~4(y)(-1+A4-1A1)+∫∂ΩH3(y-x)dσyf~3(y)(-1+A3-1A1)-∫∂ΩH2(y-x)dσyf~2(y)(-1+A2-1A1),mmmmmmmmmmmmmmmmmmx∈∂Ω,(30)g~(x)=g(x)-∫∂ΩH6(y-x)dσyf5(y)(-1+A5-1A)+∫∂ΩH5(y-x)dσyf~4(y)(-1+A4-1A)-∫∂ΩH4(y-x)dσyf~3(y)(-1+A3-1A)+∫∂ΩH3(y-x)dσyf~2(y)(-1+A2-1A)-∫∂ΩH2(y-x)dσyf~1(y)(-1+A1-1A),mmmmmmmmmmmimmmmmmx∈∂Ω.


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

Theorem 12.
SupposeG(x)∈Hα(∂Ω,Cl(Vn,n))  0<α<1 and G(x) satisfies the following condition:
(31)2n-2∥1-G(x)∥(α,∂Ω)(M+1)<1,
where M is the positive constant mentioned in Lemma 3. Then (2) admits a unique solution.Proof.
Denotingw(x)=D5[u](x) then w+(x)=w-(x)A5+f5(x), x∈∂Ω. Moreover, by D5[u](∞)=0. we get the following:
(32)ω(x)={∫∂ΩH1(y-x)dσyf5(y),x∈Ω+∫∂ΩH1(y-x)dσyf5(y)A5-1,x∈Ω-,
With Ψ1(x) being as in (20), it is easy to check that
(33)D5(u-Ψ1)(x)=0,x∈Rn∖∂Ω.
Denoting Δ2u(x)-Δ2Ψ1(x)∶=φ1(x), x∈Rn∖∂Ω, and using (Δ2u)+(x)=(Δ2u)-(x)A4+f4(x), x∈∂Ω, we conclude that
(34)φ1+(x)=φ1-(x)A4+f~4(x),x∈∂Ω,
where f~4(x)∈Hβ~(∂Ω,Cl(Vn,n)), 0<β~≤1 being as in (26). By Corollary 9, it is clear that φ1(∞)=0, and we then get the following representation formula:
(35)φ1(x)={∫∂ΩH1(y-x)dσyf~4(y),x∈Ω+∫∂ΩH1(y-x)dσyf~4(y)A4-1,x∈Ω-.
Analogously, we find with Ψ2(x) from (21) that Δ2u(x)-Δ2Ψ1(x)-Δ2Ψ2(x)=0, x∈Rn∖∂Ω. Denote
(36)D3u(x)-D3Ψ1(x)-D3Ψ2(x)∶=φ2(x),hhhhhhhhhhhhhhhhhhhhhix∈Rn∖∂Ω.
Using the condition (D3u)+(x)=(D3u)-(x)A3+f3(x), x∈∂Ω, we obtain that
(37)φ2+(x)=φ2-(x)A3+f~3(x),x∈∂Ω,
where f~3(x)∈Hβ~(∂Ω,Cl(Vn,n)), 0<β~≤1 being as in (27). By Corollary 9, it is clear that φ2(∞)=0, and we then get the following representation formula:
(38)φ2(x)={∫∂ΩH1(y-x)dσyf~3(y),x∈Ω+∫∂ΩH1(y-x)dσyf~3(y)A3-1,x∈Ω-.
HereΨ3(x) being as in (22), then we have D3u(x)-D3Ψ1(x)-D3Ψ2(x)-D3Ψ3(x)=0, x∈Rn∖∂Ω. We denote
(39)Δu(x)-ΔΨ1(x)-ΔΨ2(x)-ΔΨ3(x)∶=φ3(x)
and use the condition (Δu)+(x)=(Δu)-(x)A2+f2(x), x∈∂Ω. We conclude that
(40)φ3+(x)=φ3-(x)A2+f~2(x),x∈∂Ω,
where f~2(x)∈Hβ~(∂Ω,Cl(Vn,n)), 0<β~≤1 being as in (28). Corollary 9 ensures that φ3(∞)=0; then
(41)φ3(x)={∫∂ΩH1(y-x)dσyf~2(y),x∈Ω+∫∂ΩH1(y-x)dσyf~2(y)A2-1,x∈Ω-.
Use the same way again, Ψ4(x) being as in (23), that Δu(x)-ΔΨ1(x)-ΔΨ2(x)-ΔΨ3(x)-ΔΨ4(x)=0, x∈Rn∖∂Ω. Denoting
(42)Du(x)-∑i=14‍DΨi(x)∶=φ4(x),x∈Rn∖∂Ω
and using (Du)+(x)=(Du)-(x)A1+f1(x), x∈∂Ω, we get
(43)φ4+(x)=φ4-(x)A1+f~1(x),x∈∂Ω,
where f~1(x)∈Hβ~(∂Ω,Cl(Vn,n)), 0<β~≤1 being taken from (29). By Corollary 9, it is clear that φ4(∞)=0, and we then get the following representation formula:
(44)φ4(x)={∫∂ΩH1(y-x)dσyf~1(y),x∈Ω+∫∂ΩH1(y-x)dσyf~1(y)A1-1,x∈Ω-.
Finally, we useΨ5(x) as defined in (24) and get Du(x)-∑i=14‍DΨi(x)=0, x∈Rn∖∂Ω. Define
(45)u(x)-∑i=15‍Ψi(x)∶=φ5(x),x∈Rn∖∂Ω.
Working with the condition u+(x)=u-(x)G(x)+g(x) we arrive at
(46)φ5+(x)=φ5-(x)G(x)+g~(x),x∈∂Ω,
where g~(x)∈Hβ~(∂Ω,Cl(Vn,n)), 0<β~≤1 being taken from (30). It is clear that φ5(∞)=0. We obtain that
(47)Dφ5=0,Rn∖∂Ωφ5+(x)=φ5-(x)G(x)+g~(x),x∈∂Ωφ5(∞)=0.
We only need to consider the existence of solutions to (47). The solution to this problem may be written in the form
(48)φ5(x)=∫∂Ω‍H1(y-x)dσyφ6(y),
where φ6(y) is a Hölder continuous function to be determined on ∂Ω. Then, by using Plemelj formula, (47) can be reduced to an equivalent singular integral equation for φ6,
(49)φ6(x)=[φ6(x)2-∫∂Ω‍H1(y-x)dσyφ6(y)](1-G(x))+g~(x),mmmmmmmmmmmmmmmmmmmmmmmmmmx∈∂Ω.
LettingT denote the integral operator defined by the right hand side of (49), we get
(50)(Tφ6)(x)=[φ6(x)-(S∂Ωφ6)(x)](1-G(x))2+g(x)~.
For any ω1,ω2∈Hα(∂Ω,Cl(Vn,n)), we have
(51)∥Tω1-Tω2∥(α,∂Ω)≤2n-12∥ω1-ω2∥(α,∂Ω)∥1-G∥(α,∂Ω)(1+M).
Under the condition (31), the integral operator T is a contraction operator mapping the Banach space Hα(∂Ω,Cl(Vn,n)) into itself, which has a unique fixed point for the operator T. Thus, there exists a unique solution to (47). The proof is done.



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