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Three-dimensional Laplace cluster expansion error and work

Statement

For the single-cluster evaluator of Single-cluster three-dimensional Laplace multipole evaluator, let $Q = \sum_{j \in J} ∣ q_{j} ∣$ . If a target satisfies $r = ∣ x - c ∣ > a$ , then its potential error obeys $∣ U (x) - {\tilde{U}}_{p} (x) ∣ \leq \frac{Q}{r} \frac{(a / r)^{p + 1}}{1 - a / r} .$ If a target region is contained in a ball of radius $b$ centered at $d$ , with $R = ∣ d - c ∣ > a + b$ , then every target in the region has error at most $\frac{Q}{R - b} \frac{(a / (R - b))^{p + 1}}{1 - a / (R - b)} .$ Let $S = ∣ J ∣$ and let $T_{0}$ be the number of targets. Under the monomial-coefficient evaluation convention of the algorithm, forming moments and evaluating the truncated field costs $O ((S + T_{0}) (p + 1)^{4})$ arithmetic operations after the basis coefficient arrays have been prepared. If all basis values at sources and targets are supplied, the coefficient accumulation and expansion evaluation alone cost $O ((S + T_{0}) (p + 1)^{2})$ . This is a single-cluster estimate; it is not a 3D multilevel FMM tree complexity theorem.

Facts & Assumptions

Given: A finite 3D source cluster in a ball $B (c, a)$ , an orthonormal basis of each $H_{ℓ}$ through degree $p$ , and targets strictly outside $B (c, a)$ .

[F1]

Degree-indexed spherical harmonic spaces in three dimensions: $H_{ℓ}$ is the degree- $ℓ$ homogeneous harmonic polynomial space restricted to the unit sphere, with the surface-area inner product; its regular solid harmonics are homogeneous degree- $ℓ$ polynomials.

[F2]

Three-dimensional Laplace spherical-harmonic addition theorem: the normalized spherical-harmonic addition identity and $∣ P_{ℓ} (t) ∣ \leq 1$ hold, and $\sum_{ℓ \geq 0} r^{ℓ} P_{ℓ} (t) = (1 - 2 r t + r^{2})^{- 1 / 2}$ for $0 \leq r < 1$ .

[F3]

Single-cluster three-dimensional Laplace multipole evaluator: moments are $M_{ℓ m} = \sum_{j} q_{j} \overline{H_{ℓ m} (y_{j} - c)}$ , and the output sums the corresponding basis expansion through degree $p$ .

Proof

1.1givenF1F2algebra

For a source $y_{j}$ , put $v_{j} = y_{j} - c$ , $r_{j} = ∣ v_{j} ∣ \leq a$ , and $ω_{j} = v_{j} / r_{j}$ when $r_{j} > 0$ . For a target put $r = ∣ x - c ∣ > a$ and $ω = (x - c) / r$ . Applying the generating identity and then the addition identity in [F2] with ratio $r_{j} / r < 1$ gives $\frac{1}{∣ x - y_{j} ∣} = \sum_{ℓ = 0}^{\infty} \frac{r_{j}^{ℓ}}{r^{ℓ + 1}} P_{ℓ} (ω \cdot ω_{j}) = 4 π \sum_{ℓ = 0}^{\infty} \frac{1}{2 ℓ + 1} \frac{r_{j}^{ℓ}}{r^{ℓ + 1}} \sum_{m = - ℓ}^{ℓ} Y_{ℓ m} (ω) \overline{Y_{ℓ m} (ω_{j})}$ . For $r_{j} = 0$ , the same formula follows by the limit: the degree-zero term is $1 / r$ and all positive-degree regular solid harmonics vanish at the center.

1.2givenF1F2algebra

The dimension result in [F2] gives $\sum_{ℓ = 0}^{p} (2 ℓ + 1) = (p + 1)^{2}$ coefficients. At degree $ℓ$ , each homogeneous polynomial has at most $(\binom{ℓ + 2}{2})$ monomials and there are $2 ℓ + 1$ basis polynomials. Evaluating all basis values at one point by the prescribed monomial sums therefore uses at most a constant times $\sum_{ℓ = 0}^{p} (2 ℓ + 1) (\binom{ℓ + 2}{2}) = O ((p + 1)^{4})$ operations.

2.1givenF2F3step 1.1algebra

Summing the finite source set with weights $q_{j}$ and grouping the finite sums over $m$ produces exactly the moments and the truncated evaluator in [F3]. The remainder magnitude is at most $\sum_{j} ∣ q_{j} ∣ \sum_{ℓ = p + 1}^{\infty} \frac{r_{j}^{ℓ}}{r^{ℓ + 1}} ∣ P_{ℓ} (ω \cdot ω_{j}) ∣ \leq \frac{Q}{r} \sum_{ℓ = p + 1}^{\infty} (a / r)^{ℓ} = \frac{Q}{r} \frac{(a / r)^{p + 1}}{1 - a / r}$ , using [F2] and the geometric-series formula.

3.1givenstep 2.1algebra

If $∣ x - d ∣ \leq b$ , the triangle inequality gives $r = ∣ x - c ∣ \geq R - b > a$ . The function $a^{p + 1} / (r^{p + 1} (r - a))$ , equal to $r^{- 1} (a / r)^{p + 1} / (1 - a / r)$ , is nonincreasing for $r > a$ because its denominator is nondecreasing; substituting $r = R - b$ in step 2.1 gives the stated uniform target-region bound.

4.1F3step 1.2algebra∎

Repeating this basis evaluation at $S$ source points and $T_{0}$ target points costs $O ((S + T_{0}) (p + 1)^{4})$ . The remaining moment and target accumulations use $(p + 1)^{2}$ coefficients per point, hence $O ((S + T_{0}) (p + 1)^{2})$ additional operations; if basis values are supplied, this latter count is the full coefficient-stage cost.

Depends on

Used by

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Sources

Michael O’Neil, Integral Equations and Fast Algorithms, complete course lecture notes (2017) (standard reference, not scraped)