Chapter 5 Fast Multipole Methods

Computational Electromagnetics; Chapter 1 1 Chapter 5 Fast Multipole Methods 5.1 Near-field and far-field expansions Like the panel clustering, the Fast Multipole Method (FMM) is a technique for the fast computation of matrix-vector products in case of densely populated matrices of special structure. We consider the FMM in case of acoustic scattering. Solutions of the Helmholtz equation are given in terms of near-field and far-field expansions. Lemma 5.1 Expansion of an exterior solution Let z 1 lr 3 and B ρ1 (z 1 ) := {x lr 3 x z 1 < ρ 1 }. An outgoing solution u of the Helmholtz equation in lr 3 \ B ρ1 (z 1 ), given by means of the acoustic single layer potential u(x) = G(x, y) ϕ(y) dσ(y), (5.1) admits the representation u(x) = B ρ1 (z 1 ) α ν ( x z 1 x z 1 ) h(1) ν (k x z 1 ), (5.2) where α ν are homogeneous harmonic polynomials of degree ν given by α ν (ˆx) = ik +ν m= ν Yν m (ˆx) B ρ1 (z 1 ) j ν (k y z 1 ) Yν m ( y z 1 ) ϕ(y)dσ(y). (5.3) y z 1 The series converges uniformly on compact subsets of lr 3 \ B ρ1 (z 1 ). Proof: we obtain Using the representation G(x, y) = ik u(x) = = ik B ρ1 (z 1 ) +ν = ik m= ν B ρ1 (z 1 ) 2ν + 1 +ν m= ν P ν (ˆx ŷ) h (1) ν (k x ) j ν (k y ) = Yν m (ˆx) Yν m (ŷ) h (1) ν (k x ) j ν (k y ), G(x, y) ϕ(y) dσ(y) = h (1) ν (k x z 1 ) Yν m ( x z 1 x z 1 ) j ν (k y z 1 ) Yν m ( y z 1 ) ϕ(y) dσ(y) = y z 1

2 Ronald H.W. Hoppe = α ν ( x z 1 x z 1 ) h(1) ν (k x z 1 ). Lemma 5.2 Expansion of an interior solution Let z 2 lr 3 and B ρ2 (z i ) := {x lr 3 x z 2 < ρ 2 }. A solution v of the Helmholtz equation in B ρ2 (z 2 ), given by means of the acoustic single layer potential (5.1) with B ρ1 (z 1 ) replaced by B ρ2 (z 2 ), admits the representation v(x) = β ν ( x z 1 x z 1 ) j ν(k x z 2 ), (5.4) where β ν are homogeneous harmonic polynomials of degree ν given by β ν (ˆx) = ik +ν m= ν Yν m (ˆx) B ρ2 (z 2 ) h (1) ν (k y z 1 ) Yν m ( y z 1 ) ϕ(y)dσ(y).(5.5) y z 1 The series converges uniformly in B ρ2 (z 2 ). Proof: The assertion can be proved in the same way as that of Lemma 5.1. The behavior of an outgoing solution u in the far field can be described by the far-field representative. Lemma 5.3 Far-field representative Let u be an exterior solution of the Helmholtz equation as given by (5.2). Then, there holds uniformly for ẑ, where lim u(z 1 + t ẑ) ikt exp( ikt) = Ψ(ẑ) (5.6) t Ψ(ẑ) = ( i) ν α ν (ẑ) (5.7) is called the far-field representative. Proof: We refer to Thms. 2.5,2.15 in D. Colton and R. Kress; Inverse Acoustic and Electromagnetic Scattering Theory, Springer, Berlin-Heidelberg- New York, 1992. Likewise, the behavior of an interior solution v in the near field is given by means of the near-field representative.

Computational Electromagnetics; Chapter 1 3 Lemma 5.4 Near-field representative Let v be an interior solution of the Helmholtz equation as given by (5.4). Then, there holds v(x) = 1 Γ(ẑ) exp(ik(x z 2 ) ẑ) dσ(ẑ), (5.8) where Γ(ẑ) = is called the near-field representative. ( i) ν β ν (ẑ) (5.9) Proof: We have Γ(ẑ)exp(ik(x z 2 ) ẑ)dσ(ẑ) = ( i) ν β ν (ẑ)exp(ik(x z 2 ) ẑ)dσ(ẑ). b b Using the Jacobi-Anger expansion of exp(ik(x z 2 ) ẑ) as given by exp(ik(x z 2 ) ẑ) = we obtain = Since l=0 i l (2l + 1)j l (k x z 2 )P l ( x z 2 x z 2 ẑ), ( i) ν β ν (ẑ) i l (2l + 1)j l (k x z 2 )P l ( x z 2 ẑ) dσ(ẑ) = l=0 x z 2 ( i) ν i l x z 2 (2l + 1)j l (k x z 2 ) β ν (ẑ)p l ( ẑ) dσ(ẑ). l=0 x z 2 x z 2 β ν (ŷ)p l ( x z 2 ẑ) dσ(ẑ) = 2ν + 1 β ν( x z 2 x z 2 )δ νl, in the previous equality we can take advantage of which gives the assertion. β ν (ẑ) P l ( x z 2 ẑ) dσ(ẑ) = v(x), x z 2

4 Ronald H.W. Hoppe 5.2 Far-to-far, local-to-local, and far-to-local translations A powerful tool in the calculus of near- and far-field representatives is given by the translation operator: Lemma 5.5 Translation of near- and far-field representatives (i) Let Ψ 1 and Ψ 2 be far-field representatives of u with respect to z 1 and z 2. Then there holds Ψ 2 (ẑ) = exp(ik(z 2 z 1 ) ẑ) Ψ 1 (ẑ). (5.10) (ii) Likewise, if Γ 1 and Γ 2 are near-field representatives of v with respect to z 1 and z 2, we have Γ 2 (ẑ) = exp(ik(z 2 z 1 ) ẑ) Γ 1 (ẑ). (5.11) Proof: we refer to V. Rokhlin; Diagonal forms of translation operators for the Helmholtz equation in three dimensions, Appl. Comput. Harmonic Analysis 93, 1 82, 1993. Corollary 5.6 Special far-field representatives (i) The far-field representative of a point charge is given by exp(ik x y ) ) = ikh (1) 0 (k x y x y Ψ(ẑ) = ik exp(ik(z 1 y) ẑ). (ii) Let r. Then, the far-field representative of is given by exp(ik x y ) grad y r = x y exp(ik x y ) 1 (y x) r = (ik ) x y x y y x Ψ(ẑ) = k 2 exp(ik(z 1 y) ẑ) ẑ r. (iii) Assume that an outgoing solution u τ (x) = k(x, y) ϕ(y) dσ(y) τ is generated by τ Γ, ϕ C(Γ), and k(x, y) = α G(x, y) + β G(x,y). Then, n(y) its far-field representative is given by Ψ τ (ẑ) = ˆk(ẑ, y) ϕ(y) dσ(y), τ

Computational Electromagnetics; Chapter 1 5 where ˆk(ẑ, y) = 1 exp(ik(z 1 y) ẑ) (αik + βk 2 ẑ n(y)). Proof: Lemma 5.3 readily gives the far-field representative of ikh (1) 0 ( x y ). Assertion (i) is then a direct consequence of the transformation rule (5.10). The proof of (ii) follows in the same way observing grad y exp(ik x y ) x y Finally, taking into account that r = ik 2 h (1) 1 (k x y ) (x y) r x y Ψ τ (ẑ) = lim (u τ (z 1 + tẑ) exp( kt) ikt) = t = ϕ(y) ( lim k(z 1 + tẑ) exp( ikt)) dσ(y), t the proof of (iii) follows from (i) and (ii). τ The previous results allow us to perform local-to-local and far-to-far translations. Another important tool is the far-to-local translation which enables us to a far-field representative into a local near-field representative. Theorem 5.7 The far-to-local translation operator Let u be an outgoing solution of the Helmholtz equation in lr 3 \ B ρ1 (z 1 ) with the associated far-field representative Ψ. Moreover, assume B ρ2 (z 2 ) lr 3 with δ := z 2 z 1 > ρ 1 + ρ 2. In B ρ2 (z 2 ) we consider the near-field representative Γ n (ẑ) := ( i m (2m + 1) h (1) m (k z 2 z 1 ) P m ( z 2 z 1 z 2 z 1 ẑ)) }{{} =: µ n (z 2 z 1,ẑ). Ψ(ẑ). (5.12) Then, the function u n defined according to u n (x) := 1 Γ n (ẑ) exp(ik(x z 2 ) ẑ) dσ(ẑ) (5.13) converges to u uniformly in B ρ2 (z 2 ) u n (x) u(x) 0 (n ). Proof: First, we consider the special case u(x) = G(x, y) and y B ρ1 (z 1 ). In view of Corollary 5.6 (i) the far-field representative of u is given by Ψ(ẑ) = i k exp(ik(z 1 y) ẑ).

6 Ronald H.W. Hoppe Setting d := z 2 z 1, for x B ρ2 (z 2 ) we obtain u n (x) = ik Γ n (ẑ) exp(ikẑ (x z 2 )) dσ(ẑ) = = ik i m (2m + 1)h (1) m (k d ) P m (ˆd ẑ)exp(ik(x z 2 + z 1 y) ẑ)dσ(ẑ). The Jacobi-Anger expansion of exp(ikw ẑ), w := (x z 2 ) + (z 1 y) leads to u n (x) = ik i m (2m + 1)h (1) m (k d ) i l (2l + 1) j l (k w ) P l (ŵ ẑ) P m (ˆd ẑ)dσ(ẑ). l=0 Taking advantage of the orthogonality of the Legendre polynomials we finally get P l (ŵ ẑ) P m (ˆd ẑ) dσ(ẑ) = 2m + 1 δ lm P m (ŵ ˆd), u n (x) = ik ( 1) m (2m + 1) h (1) m (k d ) j m (k w ) P m (ŵ bd) ˆ = = ik (2m + 1) h (1) m (k d ) j m (k x z 1 ) P m ( ŵ ˆd) = = 4 π G n (d, w). Since d + w = x y, we know that u n (x) = G n (x, ) G(x, ) (n ) uniformly for x B ρ2 (z 2 ). The uniform convergence of grad G n to grad G follows by Theorem 2.10 in D. Colton and R. Kress; Inverse Acoustic and Electromagnetic Scattering Theory, Springer, Berlin-Heidelberg-New York, 1992. Let us finally consider u (x) = k(x, y) ϕ(y) dσ(y), where Γ, ϕ C(Γ) and the kernel k is given by k(x, y) = α G(x, y) + β G(x, y) n(y).

Computational Electromagnetics; Chapter 1 7 According to Corollary 5.8 (iii), we obtain the far-field representative Ψ (ẑ) = ˆk(ẑ, y) ϕ(y) dσ(y). The uniform convergence of G n to G and grad G n to grad G implies, that for n = [k(ρ 1 + ρ 2 )] + l: u,n (x) u (x) C z 1 z 2 ( ρ 1 + ρ 2 z 1 z 2 )l 1 ( α + n z 1 z 2 β ) ϕ,. Remark 5.10: We know from Chapter 3 that n has to be chosen sufficiently large, i.e., n > k(ρ 1 + ρ 2 ). In practice, the integrals over the unit sphere will be approximated by means of an appropriate quadrature formula: We denote by λ i, 0 i N, the zeroes of the (N +1)-st Legendre polynomial P N+1 and by a j := 2 (1 λ 2 j)(p N+1(λ j )) 2, 0 j N the weighting factors of the associated Gauss-Legendre quadrature formula. Setting θ j := arccos(λ j ), 0 j N, ψ l := 2πl 2N + 1, 0 l 2N, we define ˆξ ν = ˆξ (j,l) = cos(ψ l ) sin(θ j ) sin(ψ l ) cos(θ j ) cos(θ j ), w ν = w (j,l ) = 2π 2N + 1 a j. Then, the quadrature formula f(ˆx) dσ(ˆx) = w ν f(ˆξ) + R N (f) = ν Λ N ˆQ N (f) + R N (f), (5.14) where Λ N := {0, 1,..., N} {0, 1,..., 2N}, is exact for all polynomials up to the degree 2N. The following result shows that the convergence is not effected by the application of the quadrature formula ˆQ N. Theorem 5.10 Far-to-local translation by quadrature Under the assumptions of Theorem 5.8 let ũ n (x) be the approximation of u(x) = G(x, y) obtained by quadrature, i.e., ũ n (x) = ik ˆQ N (Γ n ( ) exp(ik(x z 2 ) ).

8 Ronald H.W. Hoppe For N = n + 1, we also have uniformly for x B ρ2 (z 2 ). Proof: and obtain: ũ n (x) u(x) (n ) As in the proof of Theorem 5.8 we set d := z 2 z 1, w : (x z 2 ) + (z 1 y) ũ n (x) u n (x) = k n i m (2m + 1) h (1) m (k d ) i l (2l + 1) j l (k w ) l=0 ( P l (ŵ ẑ)p m ( bd ˆ ẑ)dσ(ẑ) ˆQ N (P l (ŵ )P m (ˆd ))). Taking account of the facts that ˆQ N is exact for all polynomials up to degree 2N and has positive weights and that P l (t) 1, t 1, l 0, we get ũ n (x) u n (x) k (2m + 1) h (1) m (k d ) (2l + 1) j l (k w ). l=2n m In view of the asymptotics (2l + 1) j l (k w ) = we conclude that the infinite series (k w ) l 1 3... (2l 1) (1 + O(1 l )) = 1 2 ( e 2 l=2n m can be estimated by the first term (2l + 1) j l (k w ) k w ) l (1 + O( 1 l l )),. (4N 2m + 1) j 2N m (k w ), provided 2N m is sufficiently large. We thus obtain ũ n (x) u n (x) C (2m + 1) h (1) m (k d ) (4N 2m + 1) j 2N m (k w ). Moreover, since h (1) m (k d ) (m ), the sum over the spherical Hankel functions can be estimated by the last term in the sum, i.e., ũ n (x) u n (x) C(2n + 1) h (1) n (k d ) (4N 2n + 1) j 2N n (k w ).

Computational Electromagnetics; Chapter 1 9 We are interested that the error ũ n (x) u ( x) is of the same magnitude as u n (x) u ( x). Therefore, we must have j n+1 (k w ) (4N 2n + 1) j 2N n (k w ), which can be achieved for N = n + 1. 1 10 0 10 5 0 5 15 N 25 n Figure 5.1: Dependence of the relative error on n Figure 5.1 displays the dependence of the relative error ũ n (x) u(x) u(x) for n = 1, 2,..., 25, ρ 1 = ρ 2 = 1/4, z 2 z 1 = 1, and k =. We see that n > k(ρ 1 + ρ 2 ) is required and that the method breaks down for n > N.