JuliaApproximation / AlgebraicCurveOrthogonalPolynomials.jl

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2D on star like geometries #5

Open dlfivefifty opened 3 years ago

dlfivefifty commented 3 years ago

I think we can do any star like geometry as long as we can do the boundary: consider it as

p(x/z,y/z) = 1

for z = 0..1 (disk being classic example). Then we can construct OPs in 3-variables x,y,z from boundary OPs in 2 variables Y_{m,k}(x,y) as

Q_{m,k,i}(x,y,z) = P_k^(0,2m+1)(2z-1) * z^m * Y_{m,i}(x/z,y/z)

Claims to be double checked:

  1. This spans all polynomials in x,y,z mod the constraint p(x/z,y/z) =1, which is in fact a polynomial constraint: if p is degree d just multiply through by z^d.
  2. This basis is orthogonal w.r.t. \int_0^1 \int_{z B} f(x,y,z) g(x,y,z) dx dy dz

To get back to OPs in 2-variables we would then construct the connection matrix. Since Q_{m,k,i} spans all polynomials, it contains 2-variable polynomials as a sub space. We can compute this connection matrix by Lanczos (that is, multiply by x and y).

MarcoFasondini commented 3 years ago

Even for the disk, I don't see how to get the OPs in x, y from those in x, y, z:

3D_ops_disk
dlfivefifty commented 3 years ago

I was thinking this basis would contain both all polynomials and some other things. Doing 2-variable Lanczos would then find out how to get out just the OPs.