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2 1. Introduction

n
IR (n = 2 or 3) to denote the Cartesian co–ordinates of the points in the
n
n
Euclidean space IR . Furthermore, for x, y ∈ IR ,weset
n
1

x · y = x j y j and |x| =(x · x) 2
j=1
for the inner product and the Euclidean norm, respectively. We want to find
the solution u satisfying the differential equation
n 2
∂ v
−∆v := − 2 = f in Ω. (1.1.1)
∂x j
j=1
n
Here Ω ⊂ IR is a bounded, simply connected domain, and its boundary
2
Γ is sufficiently smooth, say twice continuously differentiable, i.e. Γ ∈ C .
(Later this assumption will be reduced.) As is known from classical analysis,
1
2
a classical solution v ∈ C (Ω) ∩ C (Ω) can be represented by boundary
potentials via the Green representation formula and the fundamental solution
E of (1.1.1). For the Laplacian, E(x, y) is given by

1
− log |x − y| for n =2,
E(x, y)= 2π (1.1.2)
1 1 for n =3.
4π |x−y|
The presentation of the solution reads
∂v ∂E(x, y)

v(x)= E(x, y) (y)ds y − v(y) ds y + E(x, y)f(y)dy
∂n ∂n y
y∈Γ y∈Γ Ω
(1.1.3)
for x ∈ Ω ( see Mikhlin [213, p. 220ff.]) where n y denotes the exterior normal
to Γ at y ∈ Γ, ds y the surface element or the arclength element for n =3
or 2, respectively, and
∂v
(y) := lim gradv(˜y) · n y . (1.1.4)
∂n ˜ y→y∈Γ,˜y∈Ω
The notation ∂/∂n y will be used if there could be misunderstanding due to
more variables.
In the case when f ≡ 0 in (1.1.1), one may also use the decomposition in
the following form:

v(x)= v p (x)+ u(x):= E(x, y)f(y)dy + u(x) (1.1.5)
R n
where u now solves the Laplace equation

−∆u =0 in Ω. (1.1.6)