The Boundary Element Method for the Helmholtz Equation ... - FEI VÅ B
The Boundary Element Method for the Helmholtz Equation ... - FEI VÅ B The Boundary Element Method for the Helmholtz Equation ... - FEI VÅ B
70 4 Discretization and Numerical Realizationand reuse some results. For the inner integral we haveS D (τ, x) := κ21ϕ 1 (y)4π ∥x − y∥ ds ywherewith F V= κ24πτ sτ0= κ2 sτ4π= κ24π0s τ − ss τ α2 sα 1 ss τ − slns τ1 dt ds(s − sx ) 2 + (t − t x ) 2 + u 2 xt − t x + (s − s x ) 2 + (t − t x ) 2 + u 2 xF D (s τ , α 2 ) − F D (0, α 2 ) − F D (s τ , α 1 ) + F D (0, α 1 )F D (s, α) := F V (s, α) 1 − p + s x− 1 I 12s τ s τfrom (4.25) and parameters p, q given by (4.26) and I 1 := s − p + s xlnαs − t x + (1 + α22 )(s − p) 2 + q 2 ds., α2 sdsα 1 sUsing per partes withu = lnαs − t x + (1 + α 2 )(s − p) 2 + q 2 , v ′ = s − p + s x,2u ′ =α (1 + α 2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p)(αs − t x ) (1 + α 2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p) 2 + q 2 , v = 1 2 (s − p)(s − s x)we obtainwithI 2 :=I 1 = 1 2 (s − p)(s − s x) lnαs − t x + (1 + α 2 )(s − p) 2 + q 2 − 1 2 I 2α (1 + α(s − p)(s − s x )2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p)(αs − t x ) (1 + α 2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p) 2 + q ds 2whereI 3 :== s22 − ps + I 3,(s − p) (t x − αs x ) (1 + α 2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p)(p − s x ) − q 2(1 + α 2 )(s − p) 2 + q 2 + (αs − t x ) (1 + α 2 )(s − p) 2 + q 2 ds.With the substitutions = p +qds√ sinh u,1 + α 2du =q√1 + α 2 cosh u
71proposed in [17], page 247, we obtainI 3 =q 2 √1 + α 2sinh uWith another substitution from [17]√1 + α 2 (t x − αs x ) cosh u + α(t x − αs x ) sinh u − q √ 1 + α 2q(1 + α 2 ) cosh u + αq √ 1 + α 2 sinh u + αs x − t xdu.v = tanh u 2v1 + v2, sinh u = , cosh u =2 1 − v2 1 − v 2 , dudv = 21 − v 2we havewith parametersI 3 =√ 4q21 + α 2v1 − v 2 11 − v 2 B 2 v 2 + B 1 v + B 0A 2 v 2 + A 1 v + A 0dvB 2 := 1 + α 2 (t x − αs x + q), A 2 := (1 + α 2 )q − (αs x − t x ),B 1 := 2α(t x − αs x ), A 1 := 2αq 1 + α 2 ,B 0 := 1 + α 2 (t x − αs x − q), A 0 := (1 + α 2 )q + (αs x − t x ).For the decomposition of the fractions in I 3 we get (see [17], pages 247–248)withHence,4q 2√1 + α 2v 1 B 2 v 2 + B 1 v + B 01 − v 2 1 − v 2 A 2 v 2 =+ A 1 v + A 0C 1 := 4q(t x − αs x )1 + α 2 , C 2 := −4qu 2 x.I 3 = 2q(t x − αs x ) 11 + α 2 1 − v 2 − 4qu2 xv1 − v 2 C11 − v 2 + C 2A 2 v 2 + A 1 v + A 0v 11 − v 2 A 2 v 2 dv.+ A 1 v + A 0The fractions from the remaining integral can be decomposed aswith coefficients4qu 2 v 1x1 − v 2 A 2 v 2 = D 1+ A 1 v + A 0 1 − v + D 21 + v + D 3 v + D 4A 2 v 2 + A 1 v + A 0D 1 :=1 u 2 x√ √1 + α 2 1 + α 2 + α , D 3 := (D 1 − D 2 )A 2 = 2u 2 xA 2 ,1 u 2 xD 2 := −√ √1 + α 2 1 + α 2 − α , D 4 := −(D 1 + D 2 )A 0 = √ 2u2 xαA 0.1 + α 2
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70 4 Discretization and Numerical Realizationand reuse some results. For <strong>the</strong> inner integral we haveS D (τ, x) := κ21ϕ 1 (y)4π ∥x − y∥ ds ywherewith F V= κ24πτ sτ0= κ2 sτ4π= κ24π0s τ − ss τ α2 sα 1 ss τ − slns τ1 dt ds(s − sx ) 2 + (t − t x ) 2 + u 2 xt − t x + (s − s x ) 2 + (t − t x ) 2 + u 2 xF D (s τ , α 2 ) − F D (0, α 2 ) − F D (s τ , α 1 ) + F D (0, α 1 )F D (s, α) := F V (s, α) 1 − p + s x− 1 I 12s τ s τfrom (4.25) and parameters p, q given by (4.26) and I 1 := s − p + s xlnαs − t x + (1 + α22 )(s − p) 2 + q 2 ds., α2 sdsα 1 sUsing per partes withu = lnαs − t x + (1 + α 2 )(s − p) 2 + q 2 , v ′ = s − p + s x,2u ′ =α (1 + α 2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p)(αs − t x ) (1 + α 2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p) 2 + q 2 , v = 1 2 (s − p)(s − s x)we obtainwithI 2 :=I 1 = 1 2 (s − p)(s − s x) lnαs − t x + (1 + α 2 )(s − p) 2 + q 2 − 1 2 I 2α (1 + α(s − p)(s − s x )2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p)(αs − t x ) (1 + α 2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p) 2 + q ds 2whereI 3 :== s22 − ps + I 3,(s − p) (t x − αs x ) (1 + α 2 )(s − p) 2 + q 2 + (1 + α 2 )(s − p)(p − s x ) − q 2(1 + α 2 )(s − p) 2 + q 2 + (αs − t x ) (1 + α 2 )(s − p) 2 + q 2 ds.With <strong>the</strong> substitutions = p +qds√ sinh u,1 + α 2du =q√1 + α 2 cosh u
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