Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 2, 361–377

Iterated Integral Operators in Clifford Analysis

H. Begehr

Dedicated to L. von Wolfersdorf on the occasion of his 65th birthday

Abstract. Integral representation formulas of Cauchy-Pompeiu type expressing Clifford-alge-bra-valued functions in domains of ℝ^{m} through its boundary values and its first order derivatives in form of the Dirac operator are iterated in order to get higher order Cauchy-Pompeiu formulas. In the most general representation formulas obtained the Dirac operator is replaced by products of powers of the Dirac and the Laplace operator. Boundary values of lower order operators are involved too. In particular the integral operators provide particular solutions to the inhomogeneous equations ∂^{k}w = f, Δ^{k}w = g and ∂Δ^{k}w = h. The main subject of this paper is to develop the representation formulas. Properties of the integral operators are not studied here. Keywords: Cauchy-Pompeiu representations, Dirac operator, Laplace operator, Clifford anal-

ysis

AMS subject classification: 31 B 10, 31 B 30, 30 G 30

1. Introduction

Any point x ∈ ℝ^{m} (2 ≤ m) with an orthonormal basis {e_{k} : 1 ≤ k ≤ m} is represented as x = ∑_{k=1}^{m} x_{k}e_{k}. By the convention

\begin{matrix}e_{1} & = 1 \\ e_{j}e_{k} + e_{k}e_{j} & = −2δ_{jk} (2 ≤ j, k ≤ m) \end{matrix} \}

a Clifford algebra is introduced (see [6, 9 - 12, 15]) consisting of elements

a = ∑_{A} a_{A}e_{A}

where the sum is taken over all ordered subsets A = {α_{1}, α_{2}, …, α_{k}} of {2, 3, …, m} with 2 ≤ α_{1} < α_{2} < ⋯ < α_{k} ≤ m and

e_{A} = e_{α_{1}}e_{α_{2}} ⋯ e_{α_{k}}.

Moreover, in the case A = ∅ the basis element e_{∅} is understood to be e_{1}. If the coefficients a_{A} are complex rather than real, then the respective algebra is denoted by ℂ_{m}. One denotes for a = ∑_{A} a_{A}e_{A} the complex conjugate as

\overline{a} = ∑_{A} \overline{a_{A}} \overline{e_{A}} where \{ \begin{matrix}\overline{e_{1}} = e_{1} = 1 \\ \overline{e_{k}} = −e_{k} (2 ≤ k ≤ m) \end{matrix}

H. Begehr: Freie Universität Berlin, I. Math. Inst., Arnimallee 3, D-14195 Berlin 362 H. Begehr

ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

and \overline{e_{A}e_{B}} = \overline{e_{A}} \overline{e_{B}}, and one defines a norm in ℂ_{m} by

|a| := ( ∑_{A} |a_{A}|^{2})^{\frac{1}{2}}

which via

|a|_{0} := 2^{\frac{m}{2}} |a|

becomes an algebra norm. ℝ^{m} is embedded into ℂ_{m}. In the sequal these elements are denoted by

z = ∑_{k=1}^{m} x_{k}e_{k}.

We remark that then

\begin{matrix}\overline{z} = x_{1} − ∑_{k=2}^{m} x_{k}e_{k}, & z\overline{z} = \overline{z}z = ∑_{k=1}^{m} x_{k}^{2} = |z|^{2} \\ z^{2} = x_{1}^{2} − ∑_{k=2}^{m} x_{k}^{2} + 2x_{1} ∑_{k=2}^{m} x_{k}e_{k}, & \overline{z}^{2} = x_{1}^{2} − ∑_{k=2}^{m} x_{k}^{2} − 2x_{1} ∑_{k=2}^{m} x_{k}e_{k}. \end{matrix}

The Dirac operator ∂ and its complex conjugate \overline{∂} given by

∂ = ∑_{k=1}^{m} e_{k} \frac{∂}{∂x_{k}} and \overline{∂} = \frac{∂}{∂x_{1}} − ∑_{k=2}^{m} e_{k} \frac{∂}{∂x_{k}}

corresponding to the Cauchy-Riemann and anti-Cauchy-Riemann operator

\frac{∂}{∂\overline{z}} = \frac{1}{2}(\frac{∂}{∂x} + i\frac{∂}{∂y} ) and \frac{∂}{∂z} = \frac{1}{2}(\frac{∂}{∂x} − i\frac{∂}{∂y} )

in ℂ, respectively, are divisors of the Laplace operator

Δ = ∂\overline{∂} = \overline{∂}∂ = ∑_{k=1}^{m} \frac{∂^{2}}{∂x_{k}^{2}} .

We note that for any z, or any z ≠ 0 and α ∈ ℝ ∂z = z∂ = 2 − m, ∂\overline{z} = \overline{z}∂ = m

∂|z|^{2} = |z|^{2}∂ = 2z, ∂|z|^{α} = |z|^{α}∂ = α|z|^{α−2}z
\overline{∂}|z|^{2} = |z|^{2}\overline{∂} = 2\overline{z}, \overline{∂}|z|^{α} = |z|^{α}\overline{∂} = α|z|^{α−2}\overline{z}
∂(\overline{z}|z|^{−m}) = (\overline{z}|z|^{−m})∂ = 0.

Lemma 1. The Dirac operator acting on polynomials follows the rules

\begin{matrix}∂\overline{z}^{k} = \overline{z}^{k}∂ = (m + 2(k − 1)) \overline{z}^{k−1} + (m − 2)z ∑_{ν=0}^{k−2} \overline{z}^{k−2−ν}z^{ν} for 2 ≤ k \\ ∂z^{k} = z^{k}∂ = (2 − m) ∑_{ν=0}^{k−1} \overline{z}^{k−ν−1}z^{ν} for 1 ≤ k. \end{matrix}

Iterated Integral Operators in Clifford Analysis 363

Proof. For the first formula observe

∂\overline{z}^{2} = \overline{z}^{2}∂ = [\overline{z}(\overline{z} + z)]∂ − |z|^{2}∂ = m(\overline{z} + z) + 2\overline{z} − 2z = (m + 2)\overline{z} + (m − 2)z and by induction

\overline{z}^{k+1}∂ = [\overline{z}^{k}(\overline{z} + z)]∂ − [\overline{z}^{k−1}|z|^{2}]∂

= (m + 2(k − 1))\overline{z}^{k−1}(\overline{z} + z) + (m − 2)z ∑_{ν=0}^{k−2} \overline{z}^{k−2−ν}z^{ν}(\overline{z} + z) + 2\overline{z}^{k}

− (m + 2(k − 2))\overline{z}^{k−2}|z|^{2} − (m − 2)z ∑_{ν=0}^{k−3} \overline{z}^{k−3−ν}z^{ν}|z|^{2} − 2\overline{z}^{k−2}|z|^{2}

= (m + 2k)\overline{z}^{k} + (m − 2)z[ ∑_{ν=0}^{k−2} \overline{z}^{k−1−ν}z^{ν} + ∑_{ν=0}^{k−2} \overline{z}^{k−2−ν}z^{ν+1} − ∑_{ν=0}^{k−3} \overline{z}^{k−2−ν}z^{ν+1}]

= (m + 2k)\overline{z}^{k} + (m − 2)z ∑_{ν=0}^{k−1} \overline{z}^{k−1−ν}z^{ν}.

Similarly, from

∂z = z∂ = 2 − m

and

z^{k+1}∂ = [z^{k}(\overline{z} + z)]∂ − [z^{k−1}|z|^{2}]∂

= (2 − m) ∑_{ν=0}^{k−1} \overline{z}^{k−ν−1}z^{ν}(\overline{z} + z) + 2z^{k} − (2 − m) ∑_{ν=0}^{k−2} \overline{z}^{k−ν−2}z^{ν}|z|^{2} − 2z^{k}

= (2 − m)[ ∑_{ν=0}^{k−1} \overline{z}^{k−ν}z^{ν} + ∑_{ν=0}^{k−1} \overline{z}^{k−ν−1}z^{ν+1} − ∑_{ν=0}^{k−2} ovz^{k−ν−1}z^{ν+1}]

= (2 − m) ∑_{ν=0}^{k} \overline{z}^{k−ν}z^{ν}

the second formula follows

Corollary 1. The Dirac operator satisfies

∂(\overline{z}^{k} + z^{k}) = (\overline{z}^{k} + z^{k})∂ = 2k \overline{z}^{k−1}

for 1 ≤ k.

For any integral representation the Stokes theorem is a fundamental tool. In Clifford analysis it has the following form (see [10, 12, 14, 16]).

Lemma 2. Let D be a bounded smooth domain in ℝ^{m} and f, g ∈ C(\overline{D}; ℂ_{m}) ∩ C^{1}(D; ℂ_{m}). Then

\begin{matrix}∫_{D}\{(f∂)g + f(∂g)\} dv & = ∫_{∂D} f dσ⃗g \\ ∫_{D}\{(f\overline{∂})g + f(\overline{∂}g)\} dv & = ∫_{∂D} f d\overline{σ⃗}g. \end{matrix} \}  (1.1)

Remarks. Here dv is the volume element of D while dσ⃗ := n⃗ dσ with n⃗ := ∑_{k=1}^{m} n_{k}e_{k} is the directed area element of ∂D where (n_{1}, …, n_{m}) is the outward directed normal vector on ∂D and dσ is the area element on ∂D. Moreover, d\overline{σ⃗} = \overline{n⃗} dσ. By the usual method from (1.1) the representation formulae of Cauchy-Pompeiu type

\begin{aligned}w(z) & = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ)w(ζ) − \frac{1}{ω_{m}} ∫_{D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} ∂w(ζ) dv(ζ) \\ w(z) & = \frac{1}{ω_{m}} ∫_{∂D} \frac{ζ − z}{|ζ − z|^{m}} d\overline{σ⃗(ζ)}w(ζ) − \frac{1}{ω_{m}} ∫_{D} \frac{ζ − z}{|ζ − z|^{m}} \overline{∂}w(ζ) dv(ζ) \end{aligned}  (1.2)

follow for w ∈ C(\overline{D}; ℂ_{m}) ∩ C^{1}(D; ℂ_{m}). Here ω_{m} is the area of the unit sphere in ℝ^{m}. Dual formulas are

\begin{aligned}w(z) & = \frac{1}{ω_{m}} ∫_{∂D} w(ζ) dσ⃗(ζ)\frac{\overline{ζ − z}}{|ζ − z|^{m}} − \frac{1}{ω_{m}} ∫_{D}(w(ζ)∂) \frac{\overline{ζ − z}}{|ζ − z|^{m}} dv(ζ) \\ w(z) & = \frac{1}{ω_{m}} ∫_{∂D} w(ζ) d\overline{σ⃗(ζ)}\frac{ζ − z}{|ζ − z|^{m}} − \frac{1}{ω_{m}} ∫_{D}(w(ζ)\overline{∂}) \frac{ζ − z}{|ζ − z|^{m}} dv(ζ). \end{aligned}  (1.2)′

The operator

T f(z) := −\frac{1}{ω_{m}} ∫_{D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} f(ζ) dv(ζ) (f ∈ L_{1}(\overline{D}; ℂ_{m}))

is known to provide a particular weak solution to the inhomogeneous Dirac equation

∂w = f in D

(see [14]) while the boundary integral in the first formula of (1.2), obviously, is a leftmonogenic function, i.e. a solution to the related homogeneous equation

∂w = 0 in D (and in ℝ^{m}\\overline{D} as well).

Iterating these representation formulas similarly as in [1 - 5, 7, 8] leads to higher order representation formulas. They provide general solutions to equations of the kinds

∂^{k}w = f,  Δ^{k}w = f,  ∂^{ℓ}Δ^{k}w = f.

The first kind of equation is treated in [16] in the homogeneous case. The second one is the inhomogeneous polyharmonic equation for ℂ_{m}-valued functions. A representation formula for the third kind of equation seems to be involved in general.

2. Higher order Dirac equation

Iterating the first formula of (1.2) leads to a representation of solutions to the inhomogeneous equation

∂^{k}w = f  in D  (2.1)

where D is a bounded smooth domain in ℝ^{m} and k ∈ ℕ.

Lemma 3. For z, ζ̃ ∈ D with z ≠ ζ̃ and k ∈ ℕ_{0}

ϕ_{k}(z, ζ̃) := \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) \frac{(\overline{ζ − ζ̃})(\overline{ζ − ζ̃}+ ζ − ζ̃)^{k}}{2^{k}k! |ζ − ζ̃|^{m}}  (2.2)

satisfies

\begin{aligned}ϕ_{0}(z, ζ̃) & = 0 \\ ∂_{z}ϕ_{k}(z, ζ̃) & = 0 \\ ϕ_{k}(z, ζ̃)∂_{ζ} & = −ϕ_{k−1}(z, ζ̃) for k ∈ ℕ \\ ϕ_{k}(z, ζ̃)∂k ; _{ζ} = (−1)^{k}ϕ_{0}(z, ζ̃) & = 0 for k ∈ ℕ_{0}. \end{aligned}

Proof. From (1.1) and ∂ \frac{\overline{z}}{|z|^{m}} = 0

ϕ_{0}(z, ζ) = \frac{1}{ω_{m}} ∫_{D} [(\frac{\overline{ζ − z}}{|ζ − z|^{m}} ∂_{ζ})\frac{\overline{ζ − ζ̃}}{|ζ − ζ̃|^{m}} + \frac{\overline{ζ − z}}{|ζ − z|^{m}} (∂_{ζ} \frac{\overline{ζ − ζ̃}}{|ζ − ζ̃|^{m}} )]dv(ζ) = 0

follows in the usual way by first applying (1.1) for D_{ε} = D\{|ζ − z| ≤ ε}. The second equation establishing the left-monogenicity of ϕ_{k} is obvious. For the remaining relations observe

∂ \frac{\overline{z}(z + \overline{z})^{k}}{|z|^{m}} = ∂ \frac{(2x_{1})^{k}\overline{z}}{|z|^{m}} = \frac{\overline{z}(z + \overline{z})^{k}}{|z|^{m}} ∂ = 2k \frac{\overline{z}(z + \overline{z})^{k−1}}{|z|^{m}} .

Hence,

\begin{aligned}∂^{k} \frac{\overline{z}(z + \overline{z})^{k}}{|z|^{m}} = \frac{\overline{z}(z + \overline{z})^{k}}{|z|^{m}} ∂^{k} & = 2^{k}k!\frac{\overline{z}}{|z|^{m}} \\ ∂^{k+1} \frac{\overline{z}(z + \overline{z})^{k}}{|z|^{m}} = \frac{\overline{z}(z + \overline{z})^{k}}{|z|^{m}} ∂^{k+1} & = 0. \end{aligned}

Applying the first formula in (1.2) for z, ζ̃ ∈ D with z ≠ ζ̃ gives

\frac{(\overline{z − ζ̃})(\overline{z − ζ̃}+ z − ζ̃)^{k}}{2^{k}k! |z − ζ̃|^{m}} ; = ϕ_{k}(z, ζ̃) − \frac{1}{ω_{m}} ∫_{D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} \frac{(\overline{ζ − ζ̃})(\overline{ζ − ζ̃}+ ζ − ζ̃)^{k−1}}{2^{k−1}(k − 1)! |ζ − ζ̃|^{m}} dv(ζ).  (2.3)

Using these formulas after having differentiated (2.2) leads to the last two formulas

Definition 1. Let D ⊂ ℝ^{m} be bounded and f ∈ L_{1}(\overline{D}; ℂ_{m}). Then for k ∈ ℕ

T_{k}f(z) := \frac{(−1)^{k}}{ω_{m}} ∫_{D} \frac{(\overline{ζ − z})(\overline{ζ − z} + ζ − z)^{k−1}}{2^{k−1}(k − 1)! |ζ − z|^{m}} f(ζ) dv(ζ) .

For k = 1 this operator T_{1} is the Pompeiu operator T, satisfying ∂T f = f (see [14]). Obviously, if T_{0}f := f, then ∂T_{k}f = T_{k−1}f for k ∈ ℕ.

Theorem 1. Let D ⊂ ℝ^{m} be bounded and smooth and w ∈ C^{k}(\overline{D}; ℂ_{m}) for k ∈ ℕ. Then for z ∈ D

\begin{aligned}w(z) & = ∑_{μ=0}^{k−1} \frac{(−1)^{μ}}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z})(\overline{ζ − z} + ζ − z)^{μ}}{2^{μ}μ! |ζ − z|^{m}} dσ⃗(ζ) ∂^{μ}w(ζ) \\ & + \frac{(−1)^{k}}{ω_{m}} ∫_{D} \frac{(\overline{ζ − z})(\overline{ζ − z} + ζ − z)^{k−1}}{2^{k−1}(k − 1)! |ζ − z|^{m}} ∂^{k}w(ζ) dv(ζ). \end{aligned}  (2.4)

Remarks. 1) An analogous formula is

\begin{aligned}w(z) & = ∑_{μ=0}^{k−1} \frac{(−1)^{μ}}{ω_{m}} ∫_{∂D} \frac{(ζ − z)(\overline{ζ − z} + ζ − z)^{μ}}{2^{μ}μ! |ζ − z|^{m}} d\overline{σ⃗(ζ)} \overline{∂}^{μ}w(ζ) \\ & + \frac{(−1)^{k}}{ω_{m}} ∫_{D} \frac{(ζ − z)(\overline{ζ − z} + ζ − z)^{k−1}}{2^{k−1}(k − 1)! |ζ − z|^{m}} \overline{∂}^{k}w(ζ) dv(ζ). \end{aligned}  (2.4)′

2) Denoting ρ_{μ} := ∂^{μ}w and

φ_{μ}(z) := \frac{(−1)^{μ}}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z})(\overline{ζ − z} + ζ − z)^{μ}}{2^{μ}μ! |ζ − z|^{m}} dσ⃗(ζ) ρ_{μ}(ζ)

for 0 ≤ μ ≤ k, then

w = ∑_{μ=0}^{k−1} φ_{μ} + T_{k}ρ_{k}.  (2.5)

Here the ∂^{μ}φ_{μ} are left-regular and ∂^{k}T_{k}ρ_{k} = ρ_{k}. Thus T_{k}ρ is a particular solution to ∂^{k}w = ρ while ∑_{μ=0}^{k−1} φ_{μ} is a solution to the related homogeneous equation ∂^{k}w = 0. (2.5) turns out to be the general solution to ∂^{k}w = ρ_{k}.

In order to express the φ_{μ} as some power series with left-regular coefficients one observes

(\overline{ζ − z} + ζ − z)^{μ} = ∑_{τ=0}^{μ}(−1)^{τ} (\overline{ζ} + ζ)^{μ−τ} (\overline{z} + z)^{τ} .

Therefore

\begin{aligned}φ_{μ}(z) & = ∑_{τ=0}^{μ} \frac{(−1)^{μ−τ}}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} (\overline{ζ} + ζ)^{μ−τ} dσ⃗(ζ) ρ_{μ}(ζ) (\overline{z} + z)^{τ} \\ & = ∑_{τ=0}^{μ} α_{τ,μ}(z) (\overline{z} + z)^{τ} \end{aligned}

Iterated Integral Operators in Clifford Analysis 367

where the α_{τ,μ}, obviously, are left-regular coefficients. So (2.5) has the form

w(z) = ∑_{τ=0}^{k−1} a_{τ} (z) (\overline{z} + z)^{τ} + T_{k}ρ (ρ := ∂^{k}w)

where the coefficients a_{τ} are left-regular. Observe

\begin{matrix}∂(\overline{z} + z)^{τ} & = 2τ(\overline{z} + z)^{τ−1} \\ ∂^{τ} (\overline{z} + z)^{τ} & = 2^{τ} τ! \end{matrix} \} .

Proof of Theorem 1. For k = 1 formula (2.4) coincides with the first Cauchy-Pompeiu formula (1.2). For k = 2 use this formula again to get

∂w(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) ∂w(ζ) − \frac{1}{ω_{m}} ∫_{D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} ∂^{2}w(ζ) dv(ζ).

Hence,

w(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) w(ζ)

− \frac{1}{ω_{m}} ∫_{∂D} ϕ̃_{1}(z, ζ̃) dσ⃗(ζ̃) ∂w(ζ̃)

+ \frac{1}{ω_{m}} ∫_{D} ϕ̃_{1}(z, ζ̃) ∂^{2}w(ζ̃) dv(ζ̃)

with

ϕ̃_{1}(z, ζ̃) = \frac{1}{ω_{m}} ∫_{D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} \frac{\overline{ζ̃− ζ}}{|ζ̃− ζ|^{m}} dv(ζ).

By the Pompeiu formula (1.2)

\frac{(\overline{z − ζ̃})(\overline{z − ζ̃}+ z − ζ̃)}{2 |z − ζ̃|^{m}} = ϕ_{1}(z_{1}, ζ̃) + ϕ̃_{1}(z, ζ̃)

and from the Green formula (1.1) and Lemma 3

\begin{aligned}∫_{∂D} ϕ_{1}(z, ζ̃)dσ⃗(ζ̃) ∂w(ζ̃) & = ∫_{D}[(ϕ_{1}(z, ζ̃)∂_{ζ}) ∂w(ζ̃) + ϕ_{1}(z, ζ̃)∂^{2}w(ζ̃)]dv(ζ̃) \\ & = ∫_{D} ϕ_{1}(z, ζ̃)∂^{2}w(ζ̃)dv(ζ̃). \end{aligned}

This shows

w(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ)w(ζ)

− \frac{1}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z})(\overline{ζ − z} + ζ − z)}{2 |ζ − z|^{m}} dσ⃗(ζ) ∂w(ζ)

+ \frac{1}{ω_{m}} ∫_{D} \frac{(\overline{ζ − z})(\overline{ζ − z} + ζ − z)}{2 |ζ − z|^{m}} ∂^{2}w(ζ) dv(ζ)

= φ_{0} + φ_{1} + T_{2}∂^{2}w

368 H. Begehr

where φ_{0} is left-regular, i.e. ∂φ_{0} = 0, and φ_{1} satisfies ∂^{2}φ_{1} = 0.

Assume now w ∈ C^{k−1}(\overline{D}; ℂ_{m}) is represented as

w = ∑_{μ=0}^{k−2} φ_{μ} + T_{k−1}∂^{k−1}w

where

φ_{μ}(z) = \frac{(−1)^{μ}}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z})(\overline{ζ − z} + ζ − z)^{μ}}{2^{μ}μ! |ζ − z|^{m}} dσ⃗(ζ) ∂^{μ}w(ζ).

Then applying this representation to ∂w for w ∈ C^{k}(\overline{D}; ℂ_{m})

∂w = ∑_{μ=0}^{k−2} φ̃_{μ} + T_{k−1}∂^{k}w

where

φ̃_{μ}(z) = \frac{(−1)^{μ}}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z})(\overline{ζ − z} + ζ − z)^{μ}}{2^{μ}μ! |ζ − z|^{m}} dσ⃗(ζ) ∂^{μ+1}w(ζ)

and the first formula from (1.2)

w = φ_{0} + T_{1}∂w

where

φ_{0}(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) w(ζ)

gives

w = φ_{0} + ∑_{μ=0}^{k−2} T_{1}φ̃_{μ} + T_{1}T_{k−1}∂^{k}w.

From (2.3)

T_{1}φ̃_{μ}(z)

= \frac{(−1)^{μ+1}}{ω_{m}} ∫_{∂D} \frac{1}{ω_{m}} ∫_{D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} \frac{(\overline{ζ̃− ζ})(\overline{ζ̃− ζ} + ζ̃− ζ)^{μ}}{2^{μ}μ! |ζ̃− ζ|^{m}} dv(ζ) dσ⃗(ζ̃) ∂^{μ+1}w(ζ̃)

= −\frac{1}{ω_{m}} ∫_{∂D} [\frac{(\overline{z − ζ̃})(\overline{z − ζ̃}+ z − ζ̃)^{μ+1}}{2^{μ+1}(μ + 1)! |ζ̃− z|^{m}} − ϕ_{μ+1}(z, ζ̃)]dσ⃗(ζ̃) ∂^{μ+1}w(ζ̃),

and from the first formula (1.1) and Lemma 3

∫_{∂D} ϕ_{μ+1}(z, ζ) dσ⃗(ζ) ∂^{μ+1}w(ζ) = ∫_{D} [(ϕ_{μ+1}(z, ζ) ∂_{ζ}) ∂^{μ+1}w(ζ) + ϕ_{μ+1}(z, ζ) ∂^{μ+2}w(ζ)]dv(ζ) = ∫_{D} [ϕ_{μ+1}(z, ζ) ∂^{μ+2}w(ζ) − ϕ_{μ}(z, ζ) ∂^{μ+1}w(ζ)] dv(ζ).

Hence,

∑_{μ=0}^{k−2} T_{1}φ̃_{μ} = ∑_{μ=1}^{k−1} φ_{μ} + \frac{1}{ω_{m}} ∫_{D} ϕ_{k−1}(z, ζ) ∂^{k}w(ζ) dv(ζ).

Moreover, from (2.3)

T_{1}T_{k−1}∂^{k}w(z)

= \frac{(−1)^{k}}{ω_{m}} ∫_{D} \frac{1}{ω_{m}} ∫_{D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} \frac{(\overline{ζ̃− ζ})(\overline{ζ̃− ζ} + ζ̃− ζ)^{k−2}}{2^{k−2}(k − 2)! |ζ̃− ζ|^{m}} dv(ζ) ∂^{k}w(ζ̃) dv(ζ̃)

= −\frac{1}{ω_{m}} ∫_{D} [ϕ_{k−1}(z, ζ̃) − \frac{(\overline{z − ζ̃})(\overline{z − ζ̃}+ z − ζ̃)^{k−1}}{2^{k−1}(k − 1)! |z − ζ̃|^{m}} ]∂^{k}w(ζ̃) dv(ζ̃).

This shows

w = ∑_{μ=0}^{k−1} φ_{μ} + T_{k}∂^{k}w,

i.e. (2.4)

3. Integral representations in terms of powers of the Laplacian

As the Laplace operator is the product of ∂ and \overline{∂} a representation formula in terms of the Laplacian can be obtained by iterating both formulas (1.2). In this section 2 < m is assumed. The case m = 2 is well-known.

Theorem 2. Let D ⊂ ℝ^{m} be a bounded and smooth domain and w ∈ C^{2}(\overline{D}; ℂ_{m}). Then for z ∈ D

\begin{aligned}w(z) & = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) w(ζ) \\ & − \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} d\overline{σ⃗(ζ)} ∂w(ζ) \\ & + \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2−m}}{2 − m} Δw(ζ) dv(ζ). \end{aligned}  (3.1)

Remark. Formula (3.1) has the form w = φ_{0} + φ_{1} + S_{1}Δw with a left-regular function φ_{0} and φ_{1} with left anti-regular ∂-derivative ∂φ_{1}, i.e. \overline{∂}(∂φ_{1}) = 0. S_{1} is the well-known potential operator

S_{1}ρ(z) = \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2−m}}{2 − m} ρ(ζ) dv(ζ) (ρ ∈ L_{1}(\overline{D}; ℂ_{m})).

Proof of Theorem 2. Applying the second formula (1.2) to ∂w gives

∂w(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{ζ − z}{|ζ − z|^{m}} d\overline{σ⃗(ζ)} ∂w(ζ) − \frac{1}{ω_{m}} ∫_{D} \frac{ζ − z}{|ζ − z|^{m}} Δw(ζ) dv(ζ).

370 H. Begehr

Thus together with the first formula (1.2)

w(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) w(ζ)

− \frac{1}{ω_{m}} ∫_{∂D} ψ̃_{1}(z, ζ̃) d\overline{σ⃗(ζ̃)} ∂w(ζ̃)

+ \frac{1}{ω_{m}} ∫_{D} ψ̃_{1}(z, ζ̃) Δw(ζ̃) dv(ζ̃)

with

ψ̃_{1}(z, ζ̃) := \frac{1}{ω_{m}} ∫_{D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} \frac{ζ̃− ζ}{|ζ̃− ζ|^{m}} dv(ζ)

follows. Setting

ψ_{1}(z, ζ̃) := \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) \frac{|ζ − ζ̃|^{2−m}}{2 − m} ,

then for z, ζ̃ ∈ D with z ≠ ζ̃ from (1.2)

\frac{|z − ζ̃|^{2−m}}{2 − m} = ψ_{1}(z, ζ̃) + ψ̃_{1}(z, ζ̃)

is seen. (1.1) then leads to

∫_{∂D} ψ_{1}(z, ζ̃) d\overline{σ⃗(ζ̃)} ∂w(ζ̃) = ∫_{D} [(ψ_{1}(z, ζ̃) \overline{∂_{ζ}}) ∂w(ζ̃) + ψ_{1}(z, ζ̃) Δw(ζ̃)]dv(ζ̃).

As ψ_{1}(z, ζ̃) \overline{∂_{ζ}} = ϕ_{0}(z, ζ̃) = 0 (see Lemma 3) this proves (3.1)

In order to generalize (3.1) the next lemma will be used.

Lemma 4. For 1 ≤ k and z ≠ 0

\begin{matrix}Δ^{k−1}|z|^{2k−m} & = 2^{k−1}(k − 1)! ∏_{ν=2}^{k}(2ν − m) |z|^{2−m} \\ Δ^{k}|z|^{2k−m} & = 0. \end{matrix} \}

Proof. Obviously, the formula holds for k = 1. Then arguing inductively

Δ^{k}|z|^{2(k+1)−m} = Δ^{k−1}\overline{∂}(2(k + 1) − m) |z|^{2k−m}z

= Δ^{k−1}(2(k + 1) − m) 2k |z|^{2k−m}

= 2^{k}k! ∏_{ν=2}^{k+1}(2ν − m) |z|^{2−m}.

This is the first formula for k + 1 rather than for k. Because of harmonicity of the right-hand side of the first formula the second follows immediately

Theorem 3. Let D ⊂ ℝ^{m} be a bounded smooth domain and w ∈ C^{2k}(\overline{D}; ℂ_{m}) for 1 ≤ k. Then

\begin{aligned}w(z) & = ∑_{μ=1}^{k} \{\frac{1}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z})|ζ − z|^{2(μ−1)−m}}{2^{μ−1}(μ − 1)! ∏_{ν=1}^{μ−1}(2ν − m)} dσ⃗(ζ) Δ^{μ−1}w(ζ) \\ & − \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2μ−m}}{2^{μ−1}(μ − 1)! ∏_{ν=1}^{μ}(2ν − m)} d\overline{σ⃗(ζ)} ∂Δ^{μ−1}w(ζ)\} \\ & + \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2k−m}}{2^{k−1}(k − 1)! ∏_{ν=1}^{k}(2ν − m)} Δ^{k}w(ζ) dv(ζ). \end{aligned}  (3.2)

Proof. For k = 1 this is just formula (3.1). Let now w ∈ C^{4}(\overline{D}; ℂ_{m}). Then from (3.1) applied as well to Δw as to w one has

w(z) = φ_{0}(z) + φ_{1}(z)

+ \frac{1}{ω_{m}} ∫_{∂D} Φ_{1}(z, ζ̃) dσ⃗(ζ̃) Δw(ζ̃)

− \frac{1}{ω_{m}} ∫_{∂D} Ψ_{1}(z, ζ̃) d\overline{σ⃗(ζ)} ∂Δ w(ζ̃)

+ \frac{1}{ω_{m}} ∫_{D} Ψ_{1}(z, ζ̃) Δ^{2}w(ζ̃) dv(ζ̃)

with

\begin{aligned}φ_{0}(z) & := \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) w(ζ) \\ φ_{1}(z) & := −\frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} \overline{dσ⃗(ζ)} ∂w(ζ) \\ Φ_{1}(z, ζ̃) & := \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2−m}}{2 − m} \frac{\overline{ζ̃− ζ}}{|ζ̃− ζ|^{m}} dv(ζ) \\ Ψ_{1}(z, ζ̃) & := \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2−m}}{2 − m} \frac{|ζ − ζ̃|^{2−m}}{2 − m} dv(ζ) \end{aligned}

where Φ_{1}(z, ζ̃) = \overline{∂_{ζ}}Ψ_{1}(z, ζ̃). From (1.2) observing

Δ |z|^{4−m} = \overline{∂}[(4 − m) z|z|^{2−m}] = 2(4 − m) |z|^{2−m}

it follows for z ≠ ζ̃

\frac{|z − ζ̃|^{4−m}}{2(4 − m)(2 − m)} = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) \frac{|ζ − ζ̃|^{4−m}}{2(4 − m)(2 − m)}

− \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} d\overline{σ⃗(ζ)} \frac{(ζ − ζ̃) |ζ − ζ̃|^{2−m}}{2(2 − m)}

+ Ψ_{1}(z, ζ̃)

372 H. Begehr

and by differentiating

\frac{(\overline{ζ̃− z})|ζ̃− z|^{2−m}}{2(2 − m)} = −Φ̃_{1}(z, ζ̃) \overline{∂_{ζ}} + Ψ̃_{1}(z, ζ̃) \overline{∂_{ζ}} + Φ_{1}(z, ζ̃)

where

\begin{aligned}Φ̃_{1}(z, ζ̃) & := \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} d\overline{σ⃗(ζ)} \frac{(ζ − ζ̃)|ζ − ζ̃|^{2−m}}{2(2 − m)} \\ Ψ̃_{1}(z, ζ̃) & := \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) \frac{|ζ − ζ̃|^{4−m}}{2(4 − m)(2 − m)}. \end{aligned}

Thus

w(z) = φ_{0}(z) + φ_{1}(z)

+ \frac{1}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ̃− z}) |ζ̃− z|^{2−m}}{2(2 − m)} dσ⃗(ζ̃) Δw(ζ̃)

− \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ̃− z|^{4−m}}{2(4 − m)(2 − m)} d\overline{σ⃗(ζ̃)} ∂Δw(ζ̃)

+ \frac{1}{ω_{m}} ∫_{D} \frac{|ζ̃− z|^{4−m}}{2(4 − m)(2 − m)} Δ^{2}w(ζ̃) dv(ζ̃)

because by (1.1)

∫_{∂D} [(Φ̃_{1}(z, ζ) − Ψ̃_{1}(z, ζ)) \overline{∂_{ζ}}]dσ⃗(ζ) Δw(ζ)

− ∫_{∂D} (Φ̃_{1}(z, ζ) − Ψ̃_{1}(z, ζ))d\overline{σ⃗(ζ)} ∂Δw(ζ) + ∫_{D} (Φ̃_{1}(z, ζ) − Ψ̃_{1}(z, ζ))Δ^{2}w(ζ) dv(ζ) = ∫_{D} [(Φ̃_{1}(z, ζ) − Ψ̃_{1}(z, ζ))∂_{ζ} \overline{∂_{ζ}}]Δw(ζ) dv(ζ) = 0

as again using (1.1)

(Φ̃_{1}(z, ζ̃) − Ψ̃_{1}(z, ζ̃))Δ_{ζ}

= −\frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} d\overline{σ⃗(ζ)} \frac{ζ̃− ζ}{|ζ̃− ζ|^{m}} − \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) \frac{|ζ − ζ̃|^{2−m}}{2 − m}

= \frac{1}{ω_{m}} ∫_{D} [\overline{∂_{ζ}} \frac{|ζ − z|^{2−m}}{2 − m} \frac{ζ − ζ̃}{|ζ − ζ̃|^{m}} − \frac{\overline{ζ − z}}{|ζ − z|^{m}} ∂_{ζ} \frac{|ζ − ζ̃|^{2−m}}{2 − m} ]dv(ζ)

= 0. This proves (3.2) in the case k = 2.

Iterated Integral Operators in Clifford Analysis 373

In order to prove (3.2) for any k > 1 assume it holds for k−1. Applying this formula for Δw leads to

Δw(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) Δw(ζ)

+ ∑_{μ=1}^{k−2} \frac{1}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z}) |ζ − z|^{2μ−n}}{2^{μ}μ! ∏_{ν=1}^{μ}(2ν − m)} dσ⃗(ζ) Δ^{μ+1}w(ζ)

− ∑_{μ=1}^{k−1} \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2μ−m}}{2^{μ−1}(μ − 1)! ∏_{ν=1}^{μ}(2ν − m)} d\overline{σ⃗(ζ)} ∂Δ^{μ}w(ζ)

+ \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2(k−1)−m}}{2^{k−2}(k − 2)! ∏_{ν=1}^{k−1}(2ν − m)} Δ^{k}w(ζ) dv(ζ).

Inserting this into (3.1) gives

w(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} dσ⃗(ζ) w(ζ) − \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} d\overline{σ⃗(ζ)} ∂w(ζ)

+ ∑_{μ=1}^{k−1} \{\frac{1}{ω_{m}} ∫_{∂D} Φ_{μ}(z, ζ̃) dσ⃗(ζ̃) Δ^{μ}w(ζ̃) − \frac{1}{ω_{m}} ∫_{∂D} Ψ_{μ}(z, ζ̃) d\overline{σ⃗(ζ̃)} ∂Δ^{μ}w(ζ̃)\}

+ \frac{1}{ω_{m}} ∫_{D} Ψ_{k−1}(z, ζ̃) Δ^{k}w(ζ̃) dv(ζ̃)

with

Φ_{μ+1}(z, ζ̃) := \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2−m}}{2 − m} \frac{(\overline{ζ̃− ζ}) |ζ̃− ζ|^{2μ−m}}{2^{μ}μ! ∏_{ν=1}^{μ}(2ν − m)} dv(ζ)
(0 ≤ μ ≤ k − 2)
Ψ_{μ}(z, ζ̃) := \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2−m}}{2 − m} \frac{|ζ̃− ζ|^{2μ−m}}{2^{μ−1}(μ − 1)! ∏_{ν=1}^{μ}(2ν − m)} dν(ζ)
(1 ≤ μ ≤ k − 1)

satisfying Ψ_{μ}(z, ζ̃) \overline{∂_{ζ}} = Φ_{μ}(z, ζ̃) for 1 ≤ μ ≤ k − 1. Formula (3.1) shows for μ ∈ ℕ_{0}

\frac{|ζ̃− z|^{2(μ+1)−m}}{2^{μ}μ! ∏_{ν=1}^{μ+1}(2ν − n)} = Ψ̃_{μ}(z, ζ̃) + Φ̃_{μ}(z, ζ̃) + Ψ_{μ}(z, ζ̃)

with

\begin{aligned}Ψ̃_{μ}(z, ζ) & := \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) \frac{|ζ̃− ζ|^{2(μ+1)−m}}{2^{μ}μ! ∏_{ν=1}^{μ+1}(2ν − m)} \\ Φ̃_{μ}(z, ζ) & := \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} d\overline{σ⃗(ζ)} \frac{(ζ̃− ζ)|ζ̃− ζ|^{2μ−m}}{2^{μ}μ! ∏_{ν=1}^{μ}(2ν − m)}. \end{aligned}

By differentiation for 1 ≤ μ ≤ k − 1

\frac{(\overline{ζ̃− z}) |ζ̃− z|^{2μ−m}}{2^{μ}μ! ∏_{ν=1}^{μ}(2ν − m)} = Ψ̃_{μ}(z, ζ̃)\overline{∂_{ζ}} + Φ̃_{μ}(z, ζ̃) \overline{∂_{ζ}} + Φ_{μ}(z, ζ̃)

374 H. Begehr

follows. Thus

w(z) = \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) w(ζ)

− \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} d\overline{σ⃗(ζ)} ∂w(ζ)

+ ∑_{μ=1}^{k−1} \frac{1}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z}) |ζ − z|^{2μ−m}}{2^{μ}μ! ∏_{ν=1}^{μ}(2ν − m)} dσ⃗(ζ) Δ^{μ}w(ζ)

− ∑_{μ=1}^{k−1} \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2(μ+1)−m}}{2^{μ}μ! ∏_{ν=1}^{μ+1}(2ν − m)} d\overline{σ⃗(ζ)} ∂Δ^{μ}w(ζ)

+ \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2k−m}}{2^{k−1}(k − 1)! ∏_{ν=1}^{k}(2ν − m)} Δ^{k}w(ζ) dv(ζ),

i.e. (3.2) follows observing

∑_{μ=1}^{k−1} \{ ∫_{∂D} [(Φ̃_{μ} + Ψ̃_{μ})(z, ζ) \overline{∂_{ζ}} ]dσ⃗(ζ) Δ^{μ}w(ζ)

− ∫_{∂D}(Φ̃_{μ} + Ψ̃_{μ})(z, ζ) d\overline{σ⃗(ζ)} ∂Δ^{μ}w(ζ)\} + ∫_{D}(Φ̃_{k−1} + Ψ̃_{k−1})(z, ζ) Δ^{k}w(ζ)dv(ζ)

= ∑_{μ=1}^{k−1} ∫_{D} {((Φ̃_{μ} + Ψ̃_{μ})(z, ζ) Δ_{ζ})Δ^{μ}w(ζ)

+ ((Φ̃_{μ} + Ψ̃_{μ})(z, ζ)\overline{∂_{ζ}})∂Δ^{μ}w(ζ) − ((Φ̃_{μ} + Ψ̃_{μ})(z, ζ)\overline{∂_{ζ}})∂Δ^{μ}w(ζ) − (Φ̃_{μ} + Ψ̃_{μ})(z, ζ)Δ^{μ+1}w(ζ)}dv(ζ) + ∫_{D}(Φ̃_{k−1} + Ψ_{k−1})(z, ζ) Δ^{k}w(ζ)dv(ζ)

= ∫_{D}(Φ̃_{0} + Ψ̃_{0})(z, ζ) w(ζ) dv(ζ) = 0.

Here (Φ̃_{μ} + Ψ̃_{μ})(z, ζ) Δ_{ζ} = (Φ̃_{μ−1} + Ψ̃_{μ−1})(z, ζ) and

(Φ̃_{0} + Ψ̃_{0})(z, ζ̃)

= \frac{1}{ω_{m}} ∫_{∂D} \frac{\overline{ζ − z}}{|ζ − z|^{m}} dσ⃗(ζ) \frac{|ζ̃− ζ|^{2−m}}{2 − m} + \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2−m}}{2 − m} d\overline{σ⃗(ζ)} \frac{ζ̃− ζ}{|ζ̃− ζ|^{m}}

= 0

is used and (1.1) applied

4. General representation

In the same way as (2.4) and (3.2) are proved by iteration these two representations could be used to develop a general formula. Because this seems to be involved when done at once one prefers a step by step procedure. We are not able to give the general formula but progressing as indicated one can get any kind of representation desired.

On the other hand such a representation formula is not of too much of interest. Rather than representing a function w from C^{2k+ℓ}(D; ℂ_{m}) through its derivatives of the kind Δ^{k}∂^{ℓ}w in D and its lower order derivatives Δ^{κ}∂^{λ}w (0 ≤ κ < k, 0 ≤ λ < ℓ) on ∂D one could just use Δ^{k+ \frac{ℓ}{2}} w in D and Δ^{κ}w (0 ≤ κ < k + \frac{ℓ}{2}) on ∂D for ℓ even and Δ^{k+ \frac{ℓ−1}{2}} ∂w in D and Δ^{κ}w, Δ^{κ}∂w (0 ≤ κ < k + \frac{ℓ−1}{2}) on ∂D for ℓ odd, respectively.

Consider the area integral

I_{k}(z) = \frac{1}{ω_{m}} ∫_{D} \frac{|ζ − z|^{2k−m}}{2^{k−1}(k − 1)! ∏_{ν=1}^{k}(2ν − m)}Δ^{k}w(ζ) dv(ζ)

in (3.2) and assume w ∈ C^{2k+1}(D; ℂ_{m}). By (1.2) then

\begin{aligned}I_{k}(z) & = \frac{1}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z}) |ζ − z|^{2k−m}}{2^{k}k! ∏_{ν=1}^{k}(2ν − m)} dσ⃗(ζ) Δ^{k}w(ζ) \\ & − \frac{1}{ω_{m}} ∫_{D}(\frac{\overline{ζ − z}) |ζ − z|^{2k−m}}{2^{k}k! ∏_{ν=1}^{k}(2ν − m)}∂Δ^{k}w(ζ) dv(ζ) \end{aligned}

follows when as usual first the domain D_{ε} = {ζ ∈ D : ε < |ζ − z|} is considered. Thus

\begin{aligned}w(z) & = ∑_{μ=0}^{k} \frac{1}{ω_{m}} ∫_{∂D} \frac{(\overline{ζ − z}) |ζ − z|^{2μ−m}}{2^{μ}μ! ∏_{ν=1}^{μ}(2ν − m)} dσ⃗(ζ) Δ^{μ}w(ζ) \\ & − ∑_{μ=1}^{k} \frac{1}{ω_{m}} ∫_{∂D} \frac{|ζ − z|^{2μ−m}}{2^{μ−1}(μ − 1)! ∏_{ν=1}^{μ}(2ν − m)} d\overline{σ⃗(ζ)} ∂Δ^{μ−1}w(ζ) \\ & − \frac{1}{ω_{m}} ∫_{D} \frac{(\overline{ζ − z}) |ζ − z|^{2k−m}}{2^{k}k! ∏_{ν=1}^{k}(2ν − m)} ∂Δ^{k}w(ζ) dv(ζ). \end{aligned}  (4.1)

376 H. Begehr

Denote the last area integral by I_{k1}(z). Then by (1.2) for w ∈ C^{2(k+1)}(D; ℂ_{m}) ∩ C^{2k}(\overline{D}; ℂ_{m})

I_{k1}(z) = \frac{1}{ω_{m}} ∫_{∂D} [\frac{(\overline{ζ − z})^{2}|ζ − z|^{2k−m}}{2^{k+1}(k + 1)! ∏_{ν=1}^{k}(2ν − m)}

+ \frac{|z|^{2(k+1)−m}}{2^{k+1}(k + 1)! ∏_{ν=2}^{k+1}(2ν − m)}]dσ⃗(ζ) ∂Δ^{k}w(ζ)

− \frac{1}{ω_{m}} ∫_{D} [\frac{(\overline{ζ − z})^{2}|ζ − z|^{2k−m}}{2^{k+1}(k + 1)! ∏_{ν=1}^{k}(2ν − m)}

+ \frac{|z|^{2(k+1)−m}}{2^{k+1}(k + 1)! ∏_{ν=2}^{k+1}(2ν − m)}]∂^{2}Δ^{k}w(ζ) dv(ζ).

This area integral I_{k2}(z) is for w ∈ C^{2(k+1)+1}(D; ℂ_{m}) ∩ C^{2(k+1)}(\overline{D}; ℂ_{m})

I_{k2}(z) = \frac{1}{ω_{m}} ∫_{∂D} [\frac{(\overline{ζ − z})^{3}|ζ − z|^{2k−m}}{2^{k+2}(k + 2)! ∏_{ν=1}^{k}(2ν − m)}

+ \frac{[2(\overline{ζ − z}) + (ζ + z)] |ζ − z|^{2(k+1)−m}}{2^{k+2}(k + 2)! ∏_{ν=2}^{k+1}(2ν − m)} ]dσ⃗(ζ) ∂^{2}Δ^{k}w(ζ)

− \frac{1}{ω_{m}} ∫_{D} [\frac{(\overline{ζ − z})^{3}|ζ − z|^{2k−m}}{2^{k+2}(k + 2)! ∏_{ν=1}^{k}(2ν − m)}

+ \frac{[2(\overline{ζ − z}) + (ζ − z)] |ζ − z|^{2(k+1)−m}}{2^{k+2}(k + 2)! ∏_{ν=2}^{k+1}(2ν − m)} ]∂^{3}Δ^{k}w(ζ) dv(ζ).

Assuming w ∈ C^{2(k+2)}(D; ℂ_{m}) ∩ C^{2(k+1)+1}(\overline{D}; ℂ_{m}) the area integral I_{k3}(z) here is

I_{k3}(z) = \frac{1}{ω_{m}} ∫_{∂D} [\frac{(\overline{ζ − z})^{4}|ζ − z|^{2k−m}}{2^{k+3}(k + 3)! ∏_{ν=1}^{k}(2ν − m)}

+ \frac{[3(\overline{ζ − z})^{2} + 2|ζ − z|^{2} + (ζ − z)^{2}] |ζ − z|^{2(k+1)−m}}{2^{k+3}(k + 3)! ∏_{ν=2}^{k+1}(2ν − m)}

+ \frac{|ζ − z|^{2(k+2)−m}}{2^{k+3}(k + 3)! ∏_{ν=3}^{k+2}(2ν − m)}]dσ⃗(ζ) ∂^{3}Δ^{k}w(ζ)

− \frac{1}{ω_{m}} ∫_{D} [\frac{(\overline{ζ − z})^{4}|ζ − z|^{2k−m}}{2^{k+3}(k + 3)! ∏_{ν=1}^{k}(2ν − m)}

+ \frac{[3(\overline{ζ − z})^{2} + 2|ζ − z|^{2} + (ζ − z)^{2}] |ζ − z|^{2(k+1)−m}}{2^{k+3}(k + 3)! ∏_{ν=2}^{k+1}(2ν − m)}

+ \frac{|ζ − z|^{2(k+2)−m}}{2^{k+3}(k + 3)! ∏_{ν=3}^{k+2}(2ν − m)}]∂^{4}Δ^{k}w(ζ) dv(ζ).

Iterated Integral Operators in Clifford Analysis 377

For proving these formulas Lemma 1 and Corollary 1 are useful.

A dual formula to (4.1) as analogously one to (2.4) can be given where ∂ is replaced by \overline{∂} and ζ − z by \overline{ζ − z}.

References

[1] Begehr, H.: Complex Analytic Methods for Partial Differential Equations. An introductory text. Singapore: World Scientific 1994.

[2] Begehr, H.: Iteration of Pompeiu operators. Memoirs on Diff. Equ. Math. Phys. 12 (1997), 13 – 21.

[3] Begehr, H.: Riemann-Hilbert Boundary Value Problems in ℂ^{n}. In: Part. Diff. and Int. Equ. (eds.: H. G. W. Begehr et al.). Dordrecht: Kluwer 1999, pp. 59 – 84.

[4] Begehr, H.: Systems of first order partial differential equations - a hypercomplex approach. In: Part. Diff. and Int. Equ. (eds.: H. G. W. Begehr et al.). Dordrecht: Kluwer 1999, pp. 155 – 175.

[5] Begehr, H. and A. Dzhuraev: An Introduction to Several Complex Variables and Partial Differential Equations. Harlow: Addison Wesley Longman 1997.

[6] Begehr, H. and R. P. Gilbert: Transformations, Transmutations, and Kernels Functions. Vol. I and II. Harlow: Longman 1992 and 1993.

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Received 03.06.1998