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Fast and Backward Stable Transforms Between Spherical Harmonic Expansions and Bivariate Fourier Series

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Fast and Backward Stable Transforms Between Spherical Harmonic Expansions and Bivariate Fourier Series Fast and backward stable transforms between spherical harmonic expansions and bivariate Fourier series Richard Mikael Slevinsky Department of Mathematics, University of Manitoba [email protected] OPSFA July 3, 2017 Spherical Harmonics Let µ be a positive Borel measure on D ⊂ Rn. The inner product: Z hf ; gi = f (x)g(x) dµ(x); D p 2 induces the norm kf k2 = hf ; f i and the Hilbert space L (D; dµ(x)). Let S2 ⊂ R3 denote the unit 2-sphere and let dΩ = sin θ dθ d'. Then any function f 2 L2(S2; dΩ) may be expanded in spherical harmonics: +1 +` +1 +1 X X m m X X m m f (θ; ') ∼ f` Y` (θ; ') = f` Y` (θ; '); `=0 m=−` m=−∞ `=jmj where the expansion coefficients are: Z m m f` = Y` (θ; ')f (θ; ') dΩ: 2 S 2 of 21 Spherical Harmonics Spherical harmonics are defined by: im' s m e m+jmj 1 (` − m)! m Y` (θ; ') = p i (` + ) P` (cos θ); ` 2 N0; −` ≤ m ≤ `: 2π 2 (` + m)! | {z } ~m P` (cos θ) Associated Legendre functions are defined by ultraspherical polynomials: 1 m m 1 m (m+ 2 ) P` (cos θ) = (−2) ( 2 )m sin θC`−m (cos θ): ~m The notation P` is used to denote orthonormality, and: Γ(x + n) (x) = n Γ(x) is the Pochhammer symbol for the rising factorial. 2 of 21 Spherical Harmonics Consider the Laplace{Beltrami operator on S2: 1 @ @ 1 @2 ∆θ;' = sin θ + : sin θ @θ @θ sin2 θ @'2 For ` 2 N0 and jmj ≤ `, the surface spherical harmonics are the eigenfunctions of ∆θ;': m m ∆θ;'Y` (θ; ') = −`(` + 1)Y` (θ; '): Spherical harmonics diagonalize the Laplace{Beltrami operator. 2 of 21 Real-(World) Applications Global Numerical Weather Prediction (NWP). Global climate models at the European Centre for Medium-Range Weather Forecasts use spherical harmonics to represent the world's climate with a horizontal resolution of approximately 10 km, corresponding to roughly 64 million degrees of freedom. Analysis of the Planck experiment. The European Space Agency sent the PLANCK satellite in orbit in 2013 to collect cosmic background radiation in an attempt to observe the first light of the universe. High resolution data are analyzed by spherical harmonics. Time-dependent Schr¨odingerequation in angular coordinates: d i jΨ(t)i = T^ + V^(t) jΨ(t)i; jΨ(t )i = jΨ i; ~ dt 0 0 models polarization effects. 3 of 21 Time Evolution Many real-world applications may be abstracted to semi-linear PDEs: ut = Lu + N (u; t); u(t0) = u0: where L is a linear (differential) operator, and N is not. Usually, L = ∆ or something similar and N (u; t) = u2, for example. Higher-order derivatives in time are reformulated as a system. Many algorithms exist for semi-linear time-stepping: Operator splitting preserves unitarity; Exponential integrating integrates the stiff linear term exactly; and, Implicit-explicit schemes are designed for versatility & efficiency. Problem: L is localized in momentum (coefficient) space and N is localized in physical (value) space. 4 of 21 Problem Fast transforms are required to convert between representations in momentum and physical spaces: Synthesis Convert an expansion in spherical harmonics to function values on the sphere. Analysis Convert function values on the sphere to spherical harmonic expansion coefficients. For a band-limit of ` ≤ n, the na¨ıvecost is O(n4) but it can be trivially reorganized to O(n3). The goal is a run-time of O(n2 logO(1) n). This has a rich history, including works by Driscoll and Healy, Mohlenhamp, Suda and Takami, Kunis and Potts, Rokhlin and Tygert and Tygert. Questions about current approaches: Which grids should be chosen? If tensor-product grids, should they be Gaussian or equispaced-in-angle? Is the method numerically stable? Important for time evolution. 5 of 21 Solution: Change the Problem The sphere is doubly periodic and supports a bivariate Fourier series. The FFT, DCT, and DST solve synthesis and analysis. Nonuniform variants extend this to arbitrary grids. eim' Y m(θ; ') = p P~m(cos θ) ) in longitude, we are done. ` 2π ` ~m The problem is to convert P` (cos θ) to Fourier series. (m+ 1 ) ~m m 2 Since P` (cos θ) / sin θC`−m (cos θ), ~m even-ordered P` are trigonometric polynomials in cos θ; and, ~m odd-ordered P` are trigonometric polynomials in sin θ. Conversions are not one-to-one. 6 of 21 The SH Connection Problem Definition Let fφn(x)gn≥0 be a family of orthogonal functions with respect to L2(D~; dµ~(x)); and, let f n(x)gn≥0 be another family of orthogonal functions with respect to L2(D; dµ(x)). The connection coefficients: h `; φnidµ c`;n = ; h `; `idµ allow for the expansion: 1 X φn(x) ∼ c`;n `(x): `=0 7 of 21 The SH Connection Problem Theorem Let fφn(x)gn≥0 and f n(x)gn≥0 be two families of orthonormal functions with respect to L2(D; dµ(x)). Then the connection coefficients satisfy: 1 X c`;mc`;n = δm;n: `=0 Any matrix A 2 Rm×n, m ≥ n, with orthonormal columns is well-conditioned and Moore{Penrose pseudo-invertible A+ = A>. ~m For every m, the P` (x) are a family of orthonormal functions for the same Hilbert space L2([−1; 1]; dx). 7 of 21 The SH Connection Problem Definition Let Gn denote the Givens rotation: 21 ··· 0 0 0 ··· 03 6: :: : : : :7 6: : : : : :7 6 7 60 ··· cn 0 sn ··· 07 6 7 Gn = 60 ··· 0 1 0 ··· 07 ; 6 7 60 · · · −sn 0 cn ··· 07 6 7 6: : : : :: :7 4: : : : : :5 0 ··· 0 0 0 ··· 1 where the sines and the cosines are in the intersections of the nth and n + 2nd rows and columns, embedded in the identity of a conformable size. 7 of 21 The SH Connection Problem Theorem ~m+2 ~m The connection coefficients between Pn+m+2(cos θ) and P`+m(cos θ) are: s 8 (` + 2m)! (n + m + 5 )n! > (2` + 2m + 1)(2m + 2) 2 ; for ` ≤ n; ` + n even; > 1 > (` + m + 2 )`! (n + 2m + 4)! m < s c`;n = (n + 1)(n + 2) > − ; for ` = n + 2; > > (n + 2m + 3)(n + 2m + 4) : 0; otherwise: Furthermore, the matrix of connection coefficients (m) (m) (m) (m) (m) C = G0 G1 ··· Gn−2Gn−1I(n+2)×n, where the sines and cosines are: s s (n + 1)(n + 2) (2m + 2)(2n + 2m + 5) sm = ; and cm = : n (n + 2m + 3)(n + 2m + 4) n (n + 2m + 3)(n + 2m + 4) 7 of 21 The SH Connection Problem \Proof." W.l.o.g., consider m = 0 and n = 6. 0 0:91287 0:0 0:31623 0:0 0:17593 0:0 1 B 0:0 0:83666 0:0 0:39641 0:0 0:24398 C B C B−0:40825 0:0 0:70711 0:0 0:3934 0:0 C B C (0) B 0:0 −0:54772 0:0 0:60553 0:0 0:37268 C C = B C B 0:0 0:0 −0:63246 0:0 0:5278 0:0 C B C B 0:0 0:0 0:0 −0:69007 0:0 0:46718 C B C @ 0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 21 The SH Connection Problem (0)> \Proof." Apply the Givens rotation G0 : 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 0:83666 0:0 0:39641 0:0 0:24398 C B C B0:0 0:0 0:7746 0:0 0:43095 0:0 C B C (0)> (0) B0:0 −0:54772 0:0 0:60553 0:0 0:37268 C G C = B C 0 B0:0 0:0 −0:63246 0:0 0:5278 0:0 C B C B0:0 0:0 0:0 −0:69007 0:0 0:46718 C B C @0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 21 The SH Connection Problem \Proof." And again: 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 1:0 0:0 0:0 0:0 0:0 C B C B0:0 0:0 0:7746 0:0 0:43095 0:0 C B C (0)> (0)> (0) B0:0 0:0 0:0 0:72375 0:0 0:44544 C G G C = B C 1 0 B0:0 0:0 −0:63246 0:0 0:5278 0:0 C B C B0:0 0:0 0:0 −0:69007 0:0 0:46718 C B C @0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 21 The SH Connection Problem \Proof." And again: 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 1:0 0:0 0:0 0:0 0:0 C B C B0:0 0:0 1:0 0:0 0:0 0:0 C B C (0)> (0)> (0)> (0) B0:0 0:0 0:0 0:72375 0:0 0:44544 C G G G C = B C 2 1 0 B0:0 0:0 0:0 0:0 0:68139 0:0 C B C B0:0 0:0 0:0 −0:69007 0:0 0:46718 C B C @0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 21 The SH Connection Problem \Proof." And again: 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 1:0 0:0 0:0 0:0 0:0 C B C B0:0 0:0 1:0 0:0 0:0 0:0 C B C (0)> (0)> (0)> (0)> (0) B0:0 0:0 0:0 1:0 0:0 0:0 C G G G G C = B C 3 2 1 0 B0:0 0:0 0:0 0:0 0:68139 0:0 C B C B0:0 0:0 0:0 0:0 0:0 0:6455 C B C @0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 21 The SH Connection Problem \Proof." And again: 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 1:0 0:0 0:0 0:0 0:0 C B C B0:0 0:0 1:0 0:0 0:0 0:0 C B C (0)> (0)> (0)> (0)> (0)> (0) B0:0 0:0 0:0 1:0 0:0 0:0 C G G G G G C = B C 4 3 2 1 0 B0:0 0:0 0:0 0:0 1:0 0:0 C B C B0:0 0:0 0:0 0:0 0:0 0:6455 C B C @0:0 0:0 0:0 0:0 0:0 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 21 The SH Connection Problem \Proof." And finally: 01:0 0:0 0:0 0:0 0:0 0:01 B0:0 1:0 0:0 0:0 0:0 0:0C B C B0:0 0:0 1:0 0:0 0:0 0:0C B C (0)> (0)> (0)> (0)> (0)> (0)> (0) B0:0 0:0 0:0 1:0 0:0 0:0C G G G G G G C = B C 5 4 3 2 1 0 B0:0 0:0 0:0 0:0 1:0 0:0C B C B0:0 0:0 0:0 0:0 0:0 1:0C B C @0:0 0:0 0:0 0:0 0:0 0:0A 0:0 0:0 0:0 0:0 0:0 0:0 7 of 21 The SH Connection Problem \Proof." Schematically: (m) C : I = : : (m) (m) G0 ··· Gn−1 Conversion between neighbouring layers is O(n) flops and storage.
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