Numerical solution of the 2+1 Teukolsky equation on a ...

Conclusion - Formulas for s ≠ 0 with ID1 and ID3· at R + :⎧⎨ l 0 + l + 3 + b −1 for l ′ = l 0µ l = l 0 + l + 4 + b −1 for l ′ = l 0 + 1⎩l ′ + l + 1 + b −1 for l ′ > l 0 + 1⎧⎨ l 0 + l + 3 −1 for l ′ = l 0· at ˜R:µ l = l 0 + l + 4 −1⎩l ′ + l + 1 −1for l ′ = l 0 + 1for l ′ > l 0 + 1· at J + :⎧l − s + 2 −1 for l ′ = l 0⎪⎨ l − s + 3 −1 for l ′ = l 0 + 1 and l = l 0µ l = l − s + 2 −1 for l ′ = l 0 + 1 and l > l 0l⎪⎩′ − s −1 for l ′ > l 0 + 1 and l < l ′ − 1l − s + 2 −1 for l ′ > l 0 + 1 and l ≥ l ′ − 1⇒ same decrement by 1 as for switch ID0 → ID218 / 20