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2.3 The Stokes Equations 69

α
i) The normal vector fields n ∈ C (Γ) where n | Γ j = 0 for = j generate
exterior to Ω on Γ = L " Γ the L–dimensional eigenspace or kernel of the
=1
1 1
simple layer operator V as well as of ( I − K ). Then the operator ( I −
2 2
K) also has an L–dimensional eigenspace generated by ϕ 0 ∈ C 1+α (Γ) with
= 0 for = j satisfying the equations
0
ϕ | Γ j
n = Dϕ 0 for = 1,L. (2.3.32)
Any eigenfunction L γ j n j generates a solution
j=1

L

0 ≡ γ j V n j and p 0 = γ 1 in Ω
j=1
(see Kohr and Pop [163], Reidinger and Steinbach [260]).
ii) On each component Γ of the boundary, the boundary integral operators
1
2
D| Γ as well as ( I + K)| Γ have the 6–dimensional eigenspace v =(a +
3
3
b × x)| Γ for all a ∈ IR with b ∈ IR .
If v j, with j =1,..., 6 and =1,...,L denotes a basis of this eigenspace
α
then there exist 6L linearly independent eigenvectors τ j, ∈ C (Γ) of the
1
adjoint operator ( I + K )| Γ ; and there holds the relation
2
τ
v j, = V | Γ j, (2.3.33)
between these two eigenspaces.
Any of the eigenfunctions v j, on Γ generates a solution

0 for x ∈ Ω if = 2,L,
u 0j, (x)= − K(x, y)v j, (x)ds x =
v j,1 (x) for x ∈ Ω if =1
Γ
⎧
⎨0 for x ∈ Ω if = 2,L,
p 0j, (x)= µ ∂
⎩ div x π(n−1) ∂n y γ n (x, y) v j,1 (y)ds y for x ∈ Ω if =1.
Γ 1
In the case n =2, the operator V needs to be replaced by

V τ := V τ + α( τds) with α> 0

Γ
an appropriately large chosen scaling constant α and a + b × x replaced by
a + b(x 2 , −x 1 ) and 6 by 3 in ii).

Proof: Let n = 3 and, for brevity, L =1.
i) It is shown by Ladyˇzenskaya in [179, p.61] that n is the only eigensolution
1

of ( I −K ). Therefore, due to the classical Fredholm alternative, the adjoint
2