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
68 4 Discretization and Numerical RealizationRemark 4.2. For integrals with other affine basis functions it is sufficient to rotate thetriangle.Remark 4.3. For x lying in the plane defined by the triangle τ (and especially in the caseof x ∈ τ) the integrand is equal to zero and we have D(τ, x) = 0.Using the local coordinates and assuming that the local coordinate vector n correspondsto the unit outward normal vector to τ, we haveD(τ, x) = 1 sτ4π 0= u sτx4π 0s τ − ss τ α2 sα 1 ss τ − ss τ1(s − s x ) 2 + u 2 xu xdt ds((t − t x ) 2 + (s − s x ) 2 + u 2 3/2x)t − t x(t − tx ) 2 + (s − s x ) 2 + u 2 x= 14πs τF K (s τ , α 2 ) − F K (0, α 2 ) − F K (s τ , α 1 ) + F K (0, α 1 ). α2 sα 1 sdsFor the function F K we have (see [17], Section C.2.2.)F K (s, α) := −αu x√ ln 1 + α 2 (s − p) + (s − s x ) 2 + (αs − t x ) 2 + u 21 + α 2 x− u x2 ln v 2 + A 1 v + B 1+ (sτ − s x ) u x|u x | arctan 2v + A 12G 1+ u x2 ln v 2 + A 2 v + B 2− (sτ − s x ) u x|u x | arctan 2v + A 2with parameters p and q defined by (4.26) andA 1 := − 2αq√ 1 + α 2u 2 x + α 2 q 2 tx − αs x1 + α 2 + q , A 2 := − 2αq√ 1 + α 2u 2 x + α 2 q 22G 2(4.27) tx − αs x1 + α 2 − q ,B 1 :=G 1 :=v :=1 + α2 tx − αs 2xu 2 x + α 2 q 2 1 + α 2 + q , B 2 := 1 + α2 tx − αs xu 2 x + α 2 q 2√1 + α 2 |u x |u 2 x + α 2 q 2 q + t √x − αs x1 + α1 + α 2 , G 2 :=2 |u x |u 2 x + α 2 q 2(1 + α 2 )(s − p) 2 + q 2 − q√ .1 + α 2 (s − p) 21 + α 2 − q ,q − t x − αs x1 + α 2 ,Singularities in the arguments of the logarithms in (4.27) and in the computation of thecoefficients A 1/2 , B 1/2 and G 1/2 can only occur when u x = 0. In such case, however, wehave D(τ, x) = 0 (see Remark 4.3). To conclude, for the situation when s = p ∧ u x ≠ 0 wehave v = 0.
694.4.5 Hypersingular Integral OperatorThe matrix corresponding to the hypersingular integral operator is given byD κ,h [i, j] := 1 4π− κ24π∂Ω∂Ω∂Ω∂Ωe iκ∥x−y∥∥x − y∥ ⟨curl ∂Ω ϕ j (y), curl ∂Ω ϕ i (x)⟩ ds y ds xe iκ∥x−y∥∥x − y∥ ϕ j(y)ϕ i (x)⟨n(x), n(y)⟩ ds y ds x .Since the unit outward normals are constant on individual elements, we have for the firstintegral D 1 1 e iκ∥x−y∥κ,h [i, j] :=4π ∂Ω ∂Ω ∥x − y∥ ⟨curl ∂Ω ϕ j (y), curl ∂Ω ϕ i (x)⟩ ds y ds x= ⟨curl ∂Ω ϕ j | τl (y), curl ∂Ω ϕ i | τk (x)⟩ 14πτ k ⊂supp ϕ i τ l ⊂supp ϕ i= ⟨curl ∂Ω ϕ j | τl (y), curl ∂Ω ϕ i | τk (x)⟩V κ,h [k, l]τ k ⊂supp ϕ i τ l ⊂supp ϕ iτ kτ le iκ∥x−y∥∥x − y∥ ds y ds xand thus the matrix entries are some linear combination of corresponding entries in V κ,h .Recall thatcurl ∂Ω ϕ(y) := n(y) × ∇ ϕ(y),where for the extensions of the basis functions we can choose functions constant along thenormal defined by the reference functionsˆϕ1 (ξ) := 1 − ξ 1 − ξ 2 , ˆϕ2 (ξ) := ξ 1 , ˆϕ3 (ξ) := ξ 2 for ξ = [ξ 1 , ξ 2 , ξ 3 ] T ∈ ˆτ × R.Hence, it is sufficient to compute the integralD 2 κ2κ,h [i, j] :=4π= κ24π+ κ24π∂Ω∂Ω∂Ωϕ i (x)ϕ i (x)ϕ i (x)∂Ω∂Ω∂Ωe iκ∥x−y∥∥x − y∥ ϕ j(y)⟨n(x), n(y)⟩ ds y ds x1∥x − y∥ ϕ j(y)⟨n(x), n(y)⟩ ds y ds xe iκ∥x−y∥ − 1∥x − y∥ ϕ j(y)⟨n(x), n(y)⟩ ds y ds x .The latter integrand is bounded and we can use the 7-point quadrature scheme. Theremaining integral is similar to the single layer potential matrix entries and although theanalytical formula is not given in [17], we will follow the computations in Section C.2.1
- Page 32 and 33: 20 2 Helmholtz Equation∂ΩΩ εε
- Page 34 and 35: 22 2 Helmholtz Equationthe normal v
- Page 36 and 37: 24 2 Helmholtz Equationand due to p
- Page 38 and 39: 26 2 Helmholtz Equationand thus l
- Page 41 and 42: 293 Boundary Integral EquationsAt t
- Page 43 and 44: 313.1 Single Layer Potential Operat
- Page 45 and 46: 33For the jump of the Neumann trace
- Page 47 and 48: 35with the surface curl operatorcur
- Page 49 and 50: 37The following theorems are standa
- Page 51 and 52: 39u ∈ H 1 (Ω, ∆+κ 2 ) if and
- Page 53 and 54: 41Another approach to computing the
- Page 55 and 56: 433.5.3 Interior Mixed Boundary Val
- Page 57 and 58: 45respectively.The unique solvabili
- Page 59: 47with an unbounded domain Ω ext :
- Page 62 and 63: 50 4 Discretization and Numerical R
- Page 64 and 65: 52 4 Discretization and Numerical R
- Page 66 and 67: 54 4 Discretization and Numerical R
- Page 68 and 69: 56 4 Discretization and Numerical R
- Page 70 and 71: 58 4 Discretization and Numerical R
- Page 72 and 73: 60 4 Discretization and Numerical R
- Page 74 and 75: 62 4 Discretization and Numerical R
- Page 76 and 77: 64 4 Discretization and Numerical R
- Page 78 and 79: 66 4 Discretization and Numerical R
- Page 82 and 83: 70 4 Discretization and Numerical R
- Page 84 and 85: 72 4 Discretization and Numerical R
- Page 87 and 88: 755 Numerical ExperimentsIn this pa
- Page 89 and 90: 77E N Err N Err N,p Err ϑ3962 1983
- Page 91 and 92: 792002000.81500.81500.60.41000.60.4
- Page 93 and 94: 81The approximate solution u h to (
- Page 95 and 96: 83with the unit direction vector d
- Page 97 and 98: 8540.540.430.430.220.320100.20.110
- Page 99: 87ConclusionIn the preceding sectio
- Page 102: 90[15] SAUTER, Stefan; SCHWAB, Chri
68 4 Discretization and Numerical RealizationRemark 4.2. For integrals with o<strong>the</strong>r affine basis functions it is sufficient to rotate <strong>the</strong>triangle.Remark 4.3. For x lying in <strong>the</strong> plane defined by <strong>the</strong> triangle τ (and especially in <strong>the</strong> caseof x ∈ τ) <strong>the</strong> integrand is equal to zero and we have D(τ, x) = 0.Using <strong>the</strong> local coordinates and assuming that <strong>the</strong> local coordinate vector n correspondsto <strong>the</strong> unit outward normal vector to τ, we haveD(τ, x) = 1 sτ4π 0= u sτx4π 0s τ − ss τ α2 sα 1 ss τ − ss τ1(s − s x ) 2 + u 2 xu xdt ds((t − t x ) 2 + (s − s x ) 2 + u 2 3/2x)t − t x(t − tx ) 2 + (s − s x ) 2 + u 2 x= 14πs τF K (s τ , α 2 ) − F K (0, α 2 ) − F K (s τ , α 1 ) + F K (0, α 1 ). α2 sα 1 sdsFor <strong>the</strong> function F K we have (see [17], Section C.2.2.)F K (s, α) := −αu x√ ln 1 + α 2 (s − p) + (s − s x ) 2 + (αs − t x ) 2 + u 21 + α 2 x− u x2 ln v 2 + A 1 v + B 1+ (sτ − s x ) u x|u x | arctan 2v + A 12G 1+ u x2 ln v 2 + A 2 v + B 2− (sτ − s x ) u x|u x | arctan 2v + A 2with parameters p and q defined by (4.26) andA 1 := − 2αq√ 1 + α 2u 2 x + α 2 q 2 tx − αs x1 + α 2 + q , A 2 := − 2αq√ 1 + α 2u 2 x + α 2 q 22G 2(4.27) tx − αs x1 + α 2 − q ,B 1 :=G 1 :=v :=1 + α2 tx − αs 2xu 2 x + α 2 q 2 1 + α 2 + q , B 2 := 1 + α2 tx − αs xu 2 x + α 2 q 2√1 + α 2 |u x |u 2 x + α 2 q 2 q + t √x − αs x1 + α1 + α 2 , G 2 :=2 |u x |u 2 x + α 2 q 2(1 + α 2 )(s − p) 2 + q 2 − q√ .1 + α 2 (s − p) 21 + α 2 − q ,q − t x − αs x1 + α 2 ,Singularities in <strong>the</strong> arguments of <strong>the</strong> logarithms in (4.27) and in <strong>the</strong> computation of <strong>the</strong>coefficients A 1/2 , B 1/2 and G 1/2 can only occur when u x = 0. In such case, however, wehave D(τ, x) = 0 (see Remark 4.3). To conclude, <strong>for</strong> <strong>the</strong> situation when s = p ∧ u x ≠ 0 wehave v = 0.
Change language