Almansi-Type Decomposition Theorem for Bi-k-regular Functions in the Clifford Algebra Cl2n+2,0(R)?

Lixia LIU Yue LIU Yonghong XIE

Abstract Almansi-type decomposition theorem for bi-k-regular functions defined in a star-like domain Ω ?Rn+1 × Rn+1 centered at the origin with values in the Clifford algebra Cl2n+2,0(R)is proved.As a corollary, Almansi-type decomposition theorem for biharmonic functions of degree k is given.

Keywords Real Clifford analysis, Biregular functions, Bi-k-regular functions,Almansi-type decomposition theorem

1 Introduction

Clifford algebra is an associative and noncommutative algebra introduced in 1878 by Clifford[2].In 1982, Brackx et al.[1] established the theoretical basis of Clifford analysis.Eriksson[3–4], Huang [6], Ren [7, 9], Sakakibara [10], Garcia [5], Qiao [8], Xie [11–12] and Yang [13]have done a lot of work in Clifford analysis.

In 2002, Malonek and Ren [7] gave the Almansi-type theorem for polymonogenic functions defined in a star-like domain ??Rnwith values in the Clifford algebraCl0,n(R).In 2006,Ren and Kahler[9]gave Almansi decomposition for polyharmonic,polyheat,and polywave functions.In 2017, Sakakibara [10] gave the method of fundamental solutions for biharmonic equation based on Almansi-type decomposition.In 2020, Garcia et al.[5] gave the decomposition of inframonogenic functions with applications in elasticity theory.

Based on the above,Almansi-type decomposition theorem for bi-k-regular functions defined in a star-like domain ??Rn+1×Rn+1centered at the origin O with values in the Clifford algebraCl2n+2,0(R)is proved.As a corollary, Almansi-type decomposition theorem for biharmonic functions of degreekin the Clifford algebraCl2n+2,0(R)is given.It generalizes the work of Reference[7] from the Clifford algebraCl0,n(R)toCl2n+2,0(R)and from one variable to two variables.

2 Preliminaries

Real Clifford algebraCln+1,0(R)is generated by {e0,e1,··· ,en}, whose identity ise?=1 and whose basise0,e1,··· ,en;e0e1,··· ,en?1en;···;e0e1···ensatisfieseiej+ejei=2δij,i,j=0,1,··· ,n,whereδijis the Kronecker sign.For any elementa ∈Cln+1,0(R),,whereA= {α1,α2,··· ,αh}, the integersαl(l= 1,2,··· ,h)satisfy 0≤α1< α2< ··· < αh ≤n,aA ∈R,eA=eα1eα2···eαhore?= 1.Define the norm of the elementa ∈Cln+1,0(R)as

Let ?0be a nonempty connected open set in Rn+1.Denote the functionf:?0→Cln+1,0(R)bywherefA ∈R.f: ?0→Cln+1,0(R)is continuous on ?0means each componentfA(x)is continuous on ?0.SupposeCr(?0,Cln+1,0(R))={f|f: ?0→Cln+1,0(R),wherefAisr-time continuously differentiable on ?0,r ∈N?}.

In this paper,we suppose that ?=?1×?2is a nonempty connected open set in Rn+1×Rn+1.

Iff ∈C2(?,Cl2n+2,0(R)), define some operators as follows:

Lemma 2.1Suppose??Rn+1×Rn+1is a domain, f,g ∈C2(?,Cl2n+2,0(R)),then for any(x,y)∈?,we have

Definition 2.1Suppose??Rn+1×Rn+1is a domain, if t(x,y)∈?holds for any(x,y)∈?and t ∈(0,1), we say that?is a star-like domain centered at the originO.

Definition 2.2Suppose??Rn+1×Rn+1is a domain, k ∈N?, f ∈C2k(?,Cl2n+2,0(R))satisfies

on?, then f is called a bi-k-regular function on?.When k=1it is called a bi-regular function on?for short.

Definition 2.3Suppose??Rn+1×Rn+1is a domain, k ∈N?, f ∈C2k(?,Cl2n+2,0(R))satisfies

on?, then f is called a bi-k-harmonic function on?.When k= 1it is called a bi-harmonic function on?for short.

3 Almansi-Type Decomposition Theorem for Bi-k-regular Functions

For anyα ≥0, we define the operatorSαby

Sα=αI+E1+E2,

whereIis a unit operator,andare Euler operators.

If ??Rn+1×Rn+1is a star-like domain centered at the origin O,β ≥1 andf ∈C2(?,Cl2n+2,0(R)), we define the operatorTβ:C2(?,Cl2n+2,0(R))→C2(?,Cl2n+2,0(R))by

Theorem 3.1If??Rn+1×Rn+1is a star-like domain centered at the originO, β ≥1and f ∈C2(?,Cl2n+2,0(R)), then for any(x,y)∈?, we have

ProofBy some straightforward calculations, we have

wheremi=txi,kj=tyj.

On the other hand,

we complete the proof.

Theorem 3.2If??Rn+1×Rn+1is a star-like domain centered at the originOand f ∈C2(?,Cl2n+2,0(R)), α1≥0,α2≥0,then for any(x,y)∈?, we have

ProofAs,we get

Thus (3.2)holds.

Theorem 3.3If??Rn+1×Rn+1is a star-like domain centered at the originO, α ≥0,f ∈C3(?,Cl2n+2,0(R)), then for any(x,y)∈?, we have

Proof

that is, (3.3)is true.Similarly, (3.4)is true.

Since

we getDx(S0f)Dy=2DxfDy+E1(DxfDy)+E2(DxfDy)=S2(DxfDy).

Similarly,

that is, (3.5)is true.

Theorem 3.4If??Rn+1×Rn+1is a star-like domain centered at the originO, β ≥1,f ∈C2(?,Cl2n+2,0(R)),then for any(x,y)∈?, we have

Proof

henceDx(Tβf)=Tβ+1(Dxf).Similarly, (Tβf)Dy=Tβ+1(fDy).

henceDx(Tβf)Dy=Tβ+2(DxfDy).

Remark 3.1If ??Rn+1×Rn+1is a star-like domain centered at the origin O,β ≥1,f ∈C3(?,Cl2n+2,0(R))is a biregular function on ?, thenSβfandTβfare biregular functions on?.

Theorem 3.5If??Rn+1×Rn+1is a star-like domain centered at the originO, f ∈C∞(?,Cl2n+2,0(R))is a biregular function on?and E2E1f=0,then for any k ∈N?, we have

ProofFirstly, we prove (3.9).

Becausefis a biregular function on ?, by Lemma 2.1 we have

Secondly, we prove (3.10).

By Lemma 2.1 we have

Hence

By (3.10)we get

Next, we prove (3.11)by induction.

Whenk=1, by (3.9)and (3.13)we have

Suppose (3.11)holds fork=l, that is,

Then whenk=l+1, by (3.2), (3.9)–(3.10)and (3.14), we have

Finally, we prove (3.12).

By (3.2), (3.10)and (3.11), we have

Remark 3.2If ??Rn+1×Rn+1is a star-like domain centered at the origin O,f ∈C∞(?,Cl2n+2,0(R))is a biregular function on ? andE2E1f=0,thenDlx(xkfyk)Dly=0(l,k ∈N?,l >k).

Let

Theorem 3.6If??Rn+1×Rn+1is a star-like domain centered at the originO, k >1,k ∈N?,f ∈C∞(?,Cl2n+2,0(R))is a bi-k-regular function on?and E2E1f= 0,then there exist uniquely f1,f2,··· ,fk where fi(i=1,2,··· ,k)satisfies DxfiDy=0on?such that

where

ProofLetGk={f ∈C∞(?,Cl2n+2,0(R)):=0}.

Step 1We prove that fork >1,k ∈N?,we have

On one hand,it is obvious thatGk?1?Gk,and it follows from Remark 3.2 that(xk?1fyk?1)=0, that is,xk?1G1yk?1?Gk.ThusGk?1+xk?1G1yk?1?Gk.

On the other hand, for anyf ∈Gk,we have

Sincef ∈Gk,we havethen,by Remark 3.1, we have.From (3.11)–(3.12), we have

Therefore, for anyk >1,Gk=Gk?1+xk?1G1yk?1.

Step 2We prove that fork >1,k ∈N?,we have

Gk=G1+xG1y+···+xk?1G1yk?1.

By Step 1, we obtain

Step 3We prove that for anyf ∈Gk,the decompositionf=g+xk?1fkyk?1(g ∈Gk?1,fk ∈G1)is unique.

Supposef=g+xk?1fkyk?1=g?+xk?1f?kyk?1, whereg,g?∈Gk?1,fk,f?k ∈G1,then

f ?f=(g ?g?)+(xk?1(fk ?f?k)yk?1)=0.

By Theorem 3.5, we have

AsTβ0=0 andSβTβ=I,fk ?f?k=0.Hence

i.e., the decompositionf=g+xk?1fkyk?1(g ∈Gk?1,fk ∈G1)is unique.

Step4By Theorem 3.5 andf=g+xk?1fkyk?1(g ∈Gk?1,fk ∈G1),we have

hence

Moreover, asg ∈Gk?1,g=g1+xk?2fk?1yk?2(g1∈Gk?2,fk?1∈G1),then by Theorem 3.5,

hence

Letg1=g2+xk?3fk?2yk?3(g2∈Gk?3,fk?2∈G1).By Theorem 3.5,

hence

In the same way, we can get the expression offk?3,fk?4,··· ,f2,f1.

Let

Remark 3.3If ??Rn+1×Rn+1is a star-like domain centered at the origin O,k ∈N?,f ∈C∞(?,Cl2n+2,0(R))is a bi-k-harmonic function on ? andE2E1f=0,then

ProofIt is obviously true fork=1.

It can be observed that

Fork=1,k ∈N?,from Theorem 3.6 it follows that

Hence