Well-conditioned boundary integral formulations for the ... - Njit

More documents

Recommendations

Info

and thus are Fredholm operators in the spaces H m− 1 2 div (Γ), m ≥ 0. If in addition η ≠ 0 and either Rη ≥ 0, Iη ≤ 0, Rζ ≤ 0, Imζ ≥ 0 or Rη ≤ 0, Iη ≥ 0, Rζ ≥ 0, Iζ ≤ 0, the integral equations (10) are uniquely solvable in the spaces H m− 1 2 div (Γ), m ≥ 0 for sufficiently large values of κ 2 . Proof. The result in equation (26) follows from those established in Lemma 2.1 and Lemma 2.2. Clearly, in the light of equation (26), the boundary integral operator that enters the CFIER formulation is Fredholm in the spaces H m− 1 2 div (Γ), m ≥ 0. Consequently, the invertibility of this operator is equivalent to its injectivity. To establish its injectivity, let a be a solution of equation (10) with E i = 0. It follows that the electromagnetic field (E s , H s ) defined by equations (8)-(9) is an outgoing solution to the Maxwell equations in the unbounded domain R 3 \D whose boundary values E s + on Γ satisfy the homogeneous conditions n × E s + = 0. In view of the uniqueness of radiating solutions for exterior Maxwell problems [29, 44] we obtain E s = H s = 0 identically in R 3 \ D. It follows then from the standard jump relations of vector layer potentials given in equations (5) that the interior traces of the electric and magnetic fields defined in formulas (8) and (9) satisfy the relations −n × E s − = a, −n × H s − = Ra, on Γ. Taking the scalar product of the second of these relations with the conjugate of the first one, using standard vector relations, integrating over Γ and appealing to the divergence theorem gives ∫ ∫ ∫ (Ra) · n × ādσ = (Ēs − × H s −) · n dσ = −ik {|H s −| 2 − |E s −| 2 }dx. (27) Γ D We have and Γ ∫ Γ Γ ∫ Γ ∫ (n × S K a) · (n × ā)dσ = Γ S K a · ādσ (28) ∫ TKdiv 1 Γ a · (n × ā)dσ = n × ∇ Γ (S K div Γ a) · (n × ā)dσ Γ∫ −−→ = − curl Γ (S K div Γ a) · (n × ā)dσ ∫Γ = − (S K div Γ a) curl Γ (n × ā)dσ ∫Γ = − (S K div Γ a) div Γ ādσ. (29) In conclusion we obtain from equations (28) and 29 that ∫ ∫ ∫ (Ra) · n × ādσ = η S K a · ādσ − ζ (S K div Γ a) div Γ ādσ. Thus, denoting by η = η R + iη I and ζ = ζ R + iζ I we get that ∫ R (Ra) · n × ādσ = η R R{S K a, a} − η I I{S K a, a} Γ Γ Γ Γ − ζ R R{S K div Γ a, div Γ a} + ζ I I{S K div Γ a, div Γ a}. (30) Given that for K = κ 1 + iκ 2 with κ ≥ 0 and ɛ ≥ 0 we have the following positivity relations which follow from equations (25) R{S K a, a} > 0, I{S K a, a} ≥ 0, R{S K div Γ a, div Γ a} ≥ 0, I{S K div Γ a, div Γ a} ≥ 0 (31) 10
for all a ∈ H −1/2 div (Γ), a ≠ 0. Taking into account the identity (30) and the positivity relations (31) we obtain that given the choice η ≠ 0 we have that if or η R ≥ 0, η I ≤ 0, ζ R ≤ 0, ζ I ≥ 0, η R ≤ 0, η I ≥ 0, ζ R ≥ 0, ζ I ≤ 0, then ∫ ∫ R (Ra) · n × ādσ > 0 or R (Ra) · n × ādσ < 0. Γ Γ The last relations and equation (27) imply that a = 0 and thus the integral operator in the left-hand side of equation (10) is injective under the assumptions in this Theorem. Remark 2.4 We note that the choice K = iκ 2 , κ 2 > 0, η = ξκ 2 , and ζ = −ξκ −1 2 , where ξ > 0 leads to regularizing operators of the form R = −ξT iκ2 that have been introduced in [30], but no rigorous proof of invertibility was provided in that reference. Also, the choice K = iκ 2 , κ 2 > 0, η = −ξκ 2 , and ζ = 0 that amounts to choosing the regularizing operator in the form R = ξ n × S iκ2 has been proposed in the literature [15]. Another choice of regularizing operators that we will consider is given by R = −γ T κ1 +iκ 2 with γ > 0, κ 1 > 0 and κ 2 > 0 large enough. In this case, η = −γ(κ 2 + iκ 1 ) and ζ = γ κ 2+iκ 1 , and thus κ 2 1 +κ2 2 the unique solvability of the CFIER equations with the choice R = −γ T κ1 +iκ 2 is not guaranteed by the result in Theorem 2.3. Nevertheless, we establish this property in the following result Theorem 2.5 If we consider the regularizing operator R = −γ T κ1 +iκ 2 κ 2 > 0 is large enough, then the boundary integral operators where γ > 0, κ 1 > 0 and have the property B k,γ,κ1 ,κ 2 ∼ + B k,γ,κ1 ,κ 2 = I 2 − K k − γT k ◦ T κ1 +iκ 2 (32) ( ) ( 1 2 + γ k 1 Π ∇Γ + 4(κ 1 + iκ 2 ) 2 + γ(κ ) 1 + iκ 2 ) 4k ( 1 2 + γ(κ 1 + iκ 2 ) 4k Π−−→ curlΓ ) Π ∇Γ ⋂ −−→ curlΓ . (33) and in addition they are continuous with bounded inverses in the spaces H m− 1 2 div (Γ), m ≥ 0. Proof. We note that relation (33) follows from the results established in Lemma 2.1 and Lemma 2.2. Just as in the proof of Theorem 2.3, the result of the Theorem is established once we show that ∫ −R (T κ1 +iκ 2 a) · n × ādσ = κ 2 R{S κ1 +iκ 2 a, a} + κ 1 I{S κ1 +iκ 2 a, a} Γ κ 2 + κ 2 1 + R{S κ1 κ2 +iκ 2 div Γ a, div Γ a} − κ 1 2 κ 2 1 + I{S κ1 κ2 +iκ 2 div Γ a, div Γ a} 2 > 0 (34) 11
Page 1 and 2: Well-conditioned boundary integral
Page 3 and 4: order effect of the pseudodifferent
Page 5 and 6: scattering problems. 2 Regularized
Page 7 and 8: In what follows we use the notation
Page 9: Since n × S k : H s (T M(Γ)) →
Page 13 and 14: Combining the estimate above with t
Page 15 and 16: We make use of the following vector
Page 17 and 18: Since γ n > 0 and RK ≥ 0 and IK
Page 19 and 20: where the operator Kk T is defined
Page 21 and 22: next the behavior of the eigenvalue
Page 23 and 24: 4 √ m 2 0 + k2 ≤ k 4/3 . It fol
Page 25 and 26: 0.62 0.6 k −1/3 max k b (k) 0.58
Page 27 and 28: Corollary 3.9 Let B = I 2 − K k
Page 29 and 30: 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 20 40
Page 31 and 32: 0.515 0.51 Coercivity constant PSB
Page 33 and 34: enclosing the scatterer; (2) the nu
Page 35 and 36: 2. Compute the surface derivatives
Page 37 and 38: D N C k A k,Sik/2 B k,2,k,0.4H 2/3
Page 39 and 40: 30 25 20 15 10 5 20 40 60 80 100 12
Page 41 and 42: Using the asymptotic expansion (For
Page 43 and 44: References [1] Abramowitz, M., and
Page 45 and 46: [32] Darbas, M., Generalized combin

Well-conditioned boundary integral formulations for the ... - Njit

Create successful ePaper yourself

Delete template?

Save as template?