Ricardo Estrada

Department of Mathematics,Louisiana State University,Baton Rouge,LA 70803,USA

Abstract.It is well-known that if p is a homogeneous polynomial of degree k in n variables,p ∈Pk,then the ordinary derivative has the form where An,k is a constant and where Y is a harmonic homogeneous polynomial of degree k,Y ∈Hk,actually the projection of p onto Hk.Here we study the distributional derivative and show that the ordinary part is still a multiple of Y,but that the delta part is independent of Y,that is,it depends only on p?Y.We also show that the exponent 2?n is special in the sense that the corresponding results for p(?)(rα)do not hold if Furthermore,we establish that harmonic polynomials appear as multiples of r2?n?2k?2k'when p(?)is applied to harmonic multipoles of the form for some

Key Words:Harmonic functions,harmonic polynomials,distributions,multipoles.

1 Introduction

It is well known[1,7,19]that any homogeneous polynomial of degree k,p ∈Pk,can be decomposed,in a unique fashion,as

where

the notation Hkbeing used to denote the harmonic homogeneous polynomials of degree k.

One can easily find the projections πk(p)and χk(p).For example,if we apply the Laplacian to(1.1)we readily obtainso that

Interestingly,these projections appear in other,somewhat surprising places.Indeed,as explained in the section Spherical Harmonics via Differentiation of[1,Chapter 5],whenever a homogeneous differential operator of degree k is applied to r2?nin Rnone obtains an expression of the form u(x)r2?n?2kwhere u is not just homogeneous of degree k,but actually belongs to Hk.In fact,more is true,since u=(2?n)(?n)···(?n?2k+4)Y,that is,if p∈Pkand we denote(2?n)(?n)···(?n?2k+4)as An,kthen

and in particular if Y∈Hkthen

Several further questions arise,however.First,since the function r2?nis singular at the origin,these formulas hold in Rn{0}but not in all Rn,so what are the corresponding formulas for the distributional derivatives?Following Farassat[6]we denote distributional derivatives with an overbar,namely,and so on.andthat is,the corresponding formulas in the whole space?Distributional derivatives of this kind play an important role in Physics;the distributional derivativeswere given by Frahm[8],and can be found in the textbooks[14]..Curiously,while in generalwill contain extra terms,namely a delta part,the distributional expressionremains basically equal to(1.5)sincedoes not have a delta part;delta parts and ordinary parts of a distribution are explained in Section 2. We give two different proofs of the formula forone by induction in Section 3 and another in Section 5.We also consider the distributional derivativein Section 4,showing that in general the ordinary part of this derivative depends only on Y,while the delta part depends only on q.

Furthermore,we show that harmonic polynomials are also obtained when we take the derivatives of multipoles§Such harmonic multipoles have received increasing attention in recent years[2];see also[18].They play a fundamental role in the ideas of the late professor Stora on convergent Feyman amplitudes[17,21].of the formfor some harmonic polynomial Y'∈Hk'.Indeed we obtain formulas for the derivativesof the principal value distribution p.v.and show that the ordinary part is a multipole of the form Z(x)/r2k'+2k+n?2for some Z∈Hk+k'.

2 Preliminaries and notation

We assume that we work in Rnwith n≥3;results in the case n=2 are usually true too,but even if true the proofs sometimes require modifications.We shall employ the notations

For results about distribution theory we refer to the textbooks [5,10,15,16,20],but give a summary of some important ideas below.The moment asymptotic expansion[5,Chapter 4]tell us that if g is a distribution defined in the whole Rnthat decays very fast at2Technically rapid decay means The expansion certainly holds if g has compact support.then g(λx)has the asymptotic expansion

where the constantsμαare the moments of g,

The notion of the finite part????Hadamard introduced the notion of the finite parts,and the name,when considering the divergent integrals that appear in of the fundamental solutions of hyperbolic equations[11].of a limit[5,Section 2.4]is as follows.Supposeis a family of strictly positive functions defined for 0<ε<ε0such that all of them tend to infinity at 0 and such that,given two different elementsthenis either 0 or ∞.

Definition 2.1.Let G(ε)be a function defined for 0<ε<ε0withThe finite part of the limit of G(ε)as ε →0+with respect to F exists and equals A if we can write??Such a decomposition is unique since any finite number of elements of F is linearly independent.G(ε)=G1(ε)+G2(ε),where G1,the infinite part,is a linear combination of the basic functions and where G2,the finite part,has the property that the limit A=limε→0+G2(ε)exists.We then employ the notation

The Hadamard finite part limit corresponds to the case whenis the family of functions ε?α|lnα|β,where α>0 and β≥0 or where α=0 and β>0.We then use the simpler notation F.p.limε→0+G(ε).

Consider now a function f defined in Rnthat is probably not integrable over the whole space but which is integrable in the region|x|>ε for any ε>0.Then the radial finite part integral??One should call the procedure(2.4)a radial finite part integral,since the use of the variable r means that f has been replaced by 0 inside a ball of radius ε.The results when solids of other shapes are removed could be very different[6,13,22].is defined as

The notation Pf(f(x))was introduced by Schwartz[20],who called it a pseudofunction.

Definition 2.2.Letbe a distribution defined in the complement of the origin.Suppose the pseudofunctionexists inLetbe any regularization of f0.Then the delta part at 0 of f is the distributionwhose support is the origin.We callthe ordinary part of f.Notice that this delta part is in fact a spherical delta part.

3 The distributional derivative

where An,k=(2?n)(?n)···(?n?2k+4).We shall give two proofs,both of which give important insight into the topic.Now we give our first proof,using induction;we give another proof in Section 5.

Let us recall[2]that if Y ∈Hkthen the pseudofunctionis actually a principal value,

If we apply the moment asymptotic expansion(2.2)to a distribution of the type g(ω)δ(r?1),wherebeing polar coordinates in Rn,we obtain the estimate.When multiplying with a harmonic polynomial the estimate is improved quite a bit.

Lemma 3.1.Let Y∈Hk.Then as ε→0+

Proof.Indeed,the expansion of Y(ω)δ(r?ε)=ε?1f(x/ε)as ε →0+,where f(x)=Y(ω)δ(r?ε),is obtained from(2.2)as

We can now give our first proof of(3.1).

Proposition 3.1.If Y∈Hkthendoes not have a delta part,that is,(3.1)holds.

Proof.If Y∈Hkthen we can write

We will prove by induction on k that λk=0 for all k.The result is of course true if k=0.Therefore,we assume that λk=0 and prove that λk+1=0.In order to do so it is enough to show that for just one harmonic polynomial Yk+1∈Hk+1we have qYk+1=0;we will do it ifwhere Y ∈Hkdepends only onThat there is at least one non zero Yk+1∈Hk+1of this form is true because n≥3.We have,

and since distributional limits and distributional derivatives can be interchanged,

4 The distributional derivative

Proposition 4.1.Let p∈Pk.Write p=Y+r2q,where Y=πk(p)∈Hk,and where q=χk(p)∈Pk?2.Then

It is useful to evaluate(4.1)at a test function.Thus if φ∈S(Rn)we obtain the formula

Notice,in particular,the extreme cases,

valid if Y∈Hk,and

that holds for all q∈Pk?2.

5 Another proof and a generalization

We now present an alternative proof of the Proposition 3.1.It is based on the formula[2,Prop.5.5]

The constant Wn,k,0equals 1 if k=0 and the product n···(n+2k?2)if k>1.

and

Comparison of the two results thus gives

giving another proof of(3.1)since(?1)k+1(2?n)Wn,k,0/(n+2k?2)equals the product(2?n)(?n)···(?n?2k+4),that is,An,k.

These ideas actually allow us to generalize(3.1).

Proposition 5.1.Ifandthen

Proof.We compute the distributional Laplacian of the convolution of the two multipolesand p.v.in two ways—using(5.1)—as we did above,equate the results,and simplify.

Notice that formula(3.1)corresponds to the case when

A different derivation of(5.2),that gives us extra information,is as follows,

We can also give a generalization of the Proposition 4.1.

Proposition 5.2.Letwithand let.Ifand q=then

Proof.Write p=Y+r2q1.Then,employing(5.3),withandand(5.1)we obtain

We have therefore encountered another instance where spherical harmonics are obtained by differentiation.Indeed,wheneverandthe ordinary derivativeis always of the formfor a harmonic polynomial Z and some constant c that depends on n,k,andWill this be the case for derivatives of the typeforThe answer is no:Just computing simple cases of this derivative for athat is not harmonic will convince the reader that if we write the result asfor somethen in general

6 The derivatives p(?)(rα)

We shall now consider how special is the exponent 2?n for obtaining spherical harmonics by differentiation.We shall see that ifand we writein the formthen in generalTo simplify our analysis we just consider the case whenbut the result remains true if α is not an even integer(a trivial case)unless α=2?n.

Let us start with the case whenHere the Funk-Hecke formula[3,(6.6)]yields,

On the other hand[3,(6.9)]

where

for some constant cα.However,the simplicity of the formula(1.4)is lost.

We finish with a generalization of the Proposition 5.1.

Proposition 6.1.If Y∈Hkandthen

Notice that,in particular,if both harmonic polynomials are of the same degree,then replacing α?2k by α we obtain