(6)
| ∇2ψ = 0, or | d2ψ | + | d2ψ | − | 1 | dψ | = 0. | |
| dx2 | dy2 | y | dy |
(7)
Changing to polar coordinates, x = r cos θ, y = r sin θ, the equation (2) becomes, with cos θ = μ,
| r2 | d2ψ | + (1 − μ2) | d2ψ | = 2 ζr3 sin θ, |
| dr2 | dμ |
(8)
of which a solution, when ζ = 0, is
| ψ = ( Arn+1 + | B | ) (1 − μ2) | dPn | = ( Arn − 1 + | B | ) y2 | dPn | , |
| rn | dμ | rn+2 | dμ |
(9)