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70 2. Boundary Integral Equations

1
operator ( I − K) has only one eigensolution ϕ , as well. It remains to show
2 0
that n is also the only linear independent eigensolution to V and satisfies
(2.3.32).
As we will show later on, for V , the Fredholm theorems are also valid and
α
V : C (Γ) → C α+1 (Γ) has the Fredholm index zero. Let τ 0 be any solution
of V τ 0 = 0. Then the single layer potential

u 0 = V τ 0 with p 0 = Φτ 0
c
is a solution of the Stokes system in Ω as well as in Ω with u 0 | Γ = 0. Then
c
3
u 0 ≡ 0 in IR and the associated pressure is zero in Ω and p 0 = β = constant
in Ω. As a consequence we have from the jump relations
!
T x (u 0− ,p 0 ) − T x (u 0+ , 0) = σ(u 0 ,p 0 )n | Γ = −βn ,
therefore τ 0 = −βn.
On the other hand, V n| Γ = 0 follows from the fact that u := V n and
p := Πn is the solution of the exterior homogeneous Neumann problem of
the Stokes system since
1

T(V u)| Γ =(− I + K )n = 0 .
2
and therefore vanishes identically (see [179, Theorem 1 p.60]).
In order to show (2.3.32), we consider the solution of the exterior Dirichlet
+
Stokes problem with u | Γ = ϕ = 0 which admits the representation
0
u(x)= Wϕ − V τ .
0
Then it follows that the corresponding simple layer term has vanishing bound-
ary values,
1
1
−V τ| Γ = ϕ − ( I + K)ϕ =( I − K)ϕ = 0 .
0 2 0 2 0
Hence, τ = βn with some constant β ∈ IR. Application of T x | Γ gives
1

τ = βn = −Dϕ +( I − K )βn = −Dϕ .
0
0
2
The case β = 0 would imply τ = 0 and then u(x) solved the homogeneous
Neumann problem which has only the trivial solution [179, p. 60] implying
ϕ = 0 which is excluded. So, β = 0 and scaling of ϕ implies (2.3.32).
0
0
1
ii) For the operator ( I+K) having the eigenspace (a+b×x)| Γ of dimension
2
1
6 we refer to [179, p. 62]. Hence, the adjoint operator ( I + K ) also has a

2
6–dimensional eigenspace due to the classical Fredholm theory since K is a
compact operator. For the operator D let us consider the potential u(x)=
c
Wv(x)in Ω with v = a+b×x| Γ . Then u is a solution of the Stokes problem
and on the boundary we find
1
+
u | Γ =( I + K)v = 0 .
2