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152 3. Representation Formulae

Applying R on both sides gives with (3.9.5)

Rγu = RγF + RC Ω (N 1 ϕ + N 2 λ);

and (3.9.23) for λ yields with (3.8.11)
Rγu = ϕ.

−1 ϕ

Hence, γu = M and the application of SM to (3.9.25) yields (3.9.24)
λ
for λ,too.
∞
iii.) If λ ∈ C (Γ) solves (3.9.24), i.e.
λ + SQ Ω P(N 1 ϕ + N 2 λ)= SγF , (3.9.26)
∞
then the potentials in (3.9.18) again define a C –solution u of the partial dif-
ferential equation (3.9.1) in Ω, and for its trace holds (3.9.25) on Γ. Applying
S yields with (3.9.26) and (3.8.11) the equation

Sγu = λ on Γ.
On the other hand, the solution u defined by (3.9.18) satisfies on Γ

Mγu = MγF + C Ω Mγu

whose second set of components reads

λ = Sγu = SγF − SQ Ω P(N 1 Rγu + N 2 λ)
in view of (3.9.23) and (3.8.11). If we subtract (3.9.26) we obtain

SC Ω N 1 (Rγu − ϕ)=0 .

On the other hand, with (3.8.28),
1
SC Ω N 1 = S{ I− DP}N 1 = −SDPN 1
2
due to (3.9.17). Therefore

SDPN 1 (ϕ − Rγu)=0

which implies ϕ = Rγu due to the assumed injectivity. Now, the first equation
(3.9.23) then follows as in ii).

Now let us consider the exterior boundary value problem for the equation
(3.6.14), i.e.,
Pu = f in Ω c (3.9.27)
together with the boundary condition