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3.9 Direct Boundary Integral Equations 155

c
Similarly, we find in Ω ,
Rγ c u = RQ Ω cPN 2 λ = −Aλ =0 and
(3.9.36)
Sγ c u = SQ Ω cPN 2 λ.
Subtracting (3.9.36) from (3.9.35) gives

Rγu = Rγ c u =0 and
Sγu − Sγ c u = S(Q Ω −Q Ω c)PN 2 λ = SP −1 PN 2 λ = λ.

This shows that λ ∈X.
ii.) Let u be the eigensolution of Pu =0 in Ω with Rγu =0 on Γ. Then u
admits the representation (3.9.18), namely

2m−1 2m− −1

u(x)= − K p P p+ +1 {N 2 Sγu} in Ω.
=0 p=0
By taking traces and applying the boundary operator R we obtain

0= Rγu = A(Sγu) .

This implies that Sγu ∈ kerA.
In the same manner we proceed for an eigensolution of the exterior prob-
lem and conclude that Sγ c u ∈ kerA as well.
This completes the proof.

Corollary 3.9.3. The injectivity of the operator

SDPN 1
in the equivalence Theorem 3.9.1 is guaranteed if and only if the interior and
exterior homogeneous boundary value problems
c
Pu = 0 in Ω and Pu c = 0 in Ω ,
Sγu = 0 and Sγ c u c = 0 on Γ with M(x; u c )=0

admit only the trivial solutions.
Proof: For the proof we exchange the rˆoles of R and S, λ and ϕ, respectively.
Then the operator A will just be SDPN 1 and Theorem 3.9.2 implies the
Corollary.

3.9.2 Transmission Problems
In this section we consider transmission problems of the following rather
general type: