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2.3 The Stokes Equations 71

c
Therefore u(x)= 0 for all x ∈ Ω and, hence,
TWv = −Dv = 0 and v ∈ kerD.
Conversely, if Dv = 0 then let u be the solution of the interior Dirichlet
problem with u | Γ = v which has the representation
−
u(x)= V τ − Wτ for x ∈ Ω
with an appropriate τ. Then applying T we find

1
1
τ =( I + K )τ + Dv =( I + K )τ .

2 2
1
Therefore τ satisfies ( I −K )τ = 0 which implies τ = βn with some β ∈ IR.

2
Hence,
u(x)= βV n(x) − Wv(x)= −Wv(x)
and its trace yields
1
( I + K)v = 0 on Γ.
2
3
Therefore v = a + b × x with some a, b ∈ IR .
1

Now let τ 0 ∈ ker( I + K ) , τ 0 = 0. Then u(x):= V τ 0 (x)in Ω is a
2
1
solution of the homogeneous Neumann problem in Ω since Tu| Γ =( I +
2
3

K )τ 0 = 0. Therefore u = a + b × x with some a, b ∈ IR and
1
V τ 0 ∈ ker( I + K) .
2
1
1

The mapping V : ker( I + K ) → ker( I + K) is also injective since for
2 2
τ 0 = 0,
V τ 0 = 0 would imply τ 0 = βn
and, hence,
1
1
1

( I + K )τ 0 = 0 = β( I + K )n = βn − ( I − K )βn = βn .
2 2 2
So, β = 0, which is a contradiction. The case L> 1 we leave to the reader
(see [143]).
For n = 2 the proof follows in the same manner and we omit the details.

In the Table 2.3.3 below we summarize the boundary integral equations
of the first and second kind for solving the four fundamental boundary value
problems together with the corresponding eigenspaces as well as the compat-
ibility conditions. We emphasize that, as a consequence of Theorem 2.3.2, the
orthogonality conditions for the right–hand side given data in the boundary
integral equations will be automatically satisfied provided the given Cauchy
data satisfy the compatibility conditions if required because of the direct
approach.
In the case of n = 2 in Table 2.3.3, replace V by V and b × x by

2

(bx 2 , −x 1 ) ,b ∈ IR , a ∈ IR .