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104Topologicalderivativeformixedsemilinearellipticproblemintwo...
25
Thesecondtermw1intheasymptoticansatz(1.38)becomesasolutionofthe
exteriorproblem:
{−∆ęw1(ę)=0,
an(ę)w1(ę)=−an(ę)ę
T∇xv(O),ę∈aω,
ę∈R2\ω,
(1.39)
Suchasolutionadmitstheasymptoticrepresentation:
w1(ę)=−
2π
1
|ę|2
ęT
m(ω)∇xv(O)+O(|ę|
−2),|ę|→∞,
wheremdenotesthevirtualmassmatrix,see[31].
Letusdenote:
F(x,V(x))=F(x,v(x)+V(x))−F(x,v(x))−V(x)F′
v(x,v(x)).
sothat:
(1.40)
FE(x;ˆ
ˆ
uE)=F(x,w(E−1x)+Ev′(x)+ˆ
uE(x))
(1.41)
+(w(E−1x)+Eao(x)+ˆ
uE(x))F′
v(x,v(x)).
Thenthethirdtermw2in(1.41)satisfiestheproblem:
{
−∆ęw2(ę)=0,
an(ę)w2(ę)=−an(ę)
1
2ęT∇xv(O)ę,ę∈aω,
ę∈R2\ω,
(1.42)
Forsuchasolution,wewritedowntheclassicalasymptoticrepresentation:
w2(ę)=
2π
c
ln
|ę|
1
+O(
|ę|
1
)|ę|→∞,
wheretheconstantcintheaboverelationcanbecalculatedasfollows:
∫
an(ę)w2(ę)dsę=−∫
a|ę|
a
2π
c
ln
|ę|
1
dę=c.
aω
aBR
BytheGreenformula,wecomputetheleftboundaryintegral:
(1.43)
−∫
an(ę)
1
2
ęT∇2
xv(O)ędsę=∫
∆ę
1
2
ęT∇2
xv(O)ędę
(1.44)
aω
ω
=mes2ω∆xv(O)=−mes2ωF(O;v(O)).
Finally,thefourthtermv′in(1.41)istobefoundfromtheDirichletproblem:
(
I
I
I
I
−∆xv′(x)=(−
2π
1
|x|2
xT
m(ω)∇xv(O)−
2π
1
ln
|x|
E
mes2ωF(O;v(O))
4
+v′(x))F′
v(x,v(x)),
x∈Ω,
I
I
I
v′(x)=
2π
1
|x|2
xT
m(ω)∇xv(O)+
2π
1
ln
|x|
E
mes2ωF(O;v(O),
I
x∈aΩ,
l
(1.45)
