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154    3. Representation Formulae


                           In comparison to equations (3.9.23) and (3.9.32), we see that A Ω and A Ω c
                           are the same for interior and exterior problems. Hence, we simply denote this
                           operator in the sequel by

                                               A := A Ω = A Ω c = −RDPN 2 .

                           It is not difficult to see that under the same assumptions as for the interior
                           problem, the equivalence Theorem 3.9.1 remains valid for the exterior
                                                                                            c
                           problem (3.9.27), (3.9.28) with the radiation condition provided f ∈ C (Ω )
                                                                                        ∞
                                                        n
                           and f has compact support in IR .
                              The following theorem shows the relation between the solutions of the ho-
                           mogeneous boundary value problems and the homogeneous boundary integral
                           equations of the first kind.
                           Theorem 3.9.2. Under the previous assumptions Γ ∈ C  ∞  and a α ∈ C ,
                                                                                           ∞
                           we have for the eigenspaces of the interior and exterior boundary value prob-
                           lems and for the eigenspace of the boundary integral operator A on Γ, respec-
                           tively, the relation:

                                            kerA = X                                   (3.9.34)
                            where

                                                           ∞
                                            kerA = {λ ∈ C (Γ)   Aλ =0 on Γ}
                           and

                                                     ∞
                                 X = span {Sγu   u ∈ C (Ω) ∧ Pu =0 in Ω and Rγu =0 on Γ}

                                                                      c
                                                  ∞
                                                      c
                                   ∪{Sγ c u c   u c ∈ C (Ω ) ∧ Pu c =0 in Ω ,Rγ c u c =0 on Γ

                                       and the radiation condition M(x; u c )=0}
                           Proof:
                              i.) Suppose λ ∈ kerA, i.e. Aλ =0 on Γ. Then define the potential
                                                   2m−1 2m− −1

                                            u(x):=            K p P p+ +1 (N 2 λ)
                                                     =0  p=0
                           where (ψ)   denotes the  –th component of the vector ψ.Now, u is a solution
                           of
                                                                       c
                                                 Pu =0    in Ω and in Ω .
                           Then we find in Ω
                                           Rγu  = RQ Ω PN 2 λ = −Aλ =0 and
                                                                                       (3.9.35)
                                           Sγu  = SQ Ω PN 2 λ.
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