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656    CHAPTER 18  The Potential Equation

                                                                    z


                                                                      ϕ
                                                                      ρ
                                                                              y
                                                                    θ

                                                              x
                                                              FIGURE 18.3 Spher-
                                                              ical coordinates.



                                    Assuming symmetry of the temperature function about the z-axis, the solution is inde-
                                 pendent of θ and depends only on ρ and ϕ. Laplace’s equation in spherical coordinates (with
                                 independence from θ)is

                                                                            2
                                                             2
                                                            ∂ u   2 ∂u   1 ∂ u  cot(ϕ) ∂u
                                                   2
                                                 ∇ u(ρ,ϕ) =     +     +       +          = 0.
                                                            ∂ρ 2  ρ ∂ρ  ρ ∂ϕ  2   ρ  2  ∂ϕ
                                                                         2
                                 The temperature on the surface is u(R,ϕ) = f (ϕ) with f being given.
                                    Let u(ρ,ϕ) = X(ρ)
(ϕ) to obtain
                                                            2       1       cot(ϕ)

                                                     X 
 +   X 
 +   X
 +        X
 = 0.
                                                            ρ      ρ  2      ρ 2
                                 Upon dividing this equation by X
, we can separate the variables, obtaining
                                                      
         
        X      X
                                                         + cot(ϕ)  =−ρ  2   − 2ρ  =−λ.
                                                       
         
       X      X
                                 Then
                                                  2


                                                 ρ X + 2ρX − λX = 0 and 
 + cot(ϕ)
 + λ
 = 0.


                                 To solve this equation for 
, write it as
                                                             1
                                                                 [
 sin(ϕ)] + λ
 = 0.                   (18.5)


                                                           sin(ϕ)
                                 Change variables by putting x = cos(ϕ). Then ϕ = arccos(x).Let
                                                              G(x) = 
(arccos(x)).

                                 Since 0 ≤ ϕ ≤ π, then −1 ≤ x ≤ 1. Compute
                                                                   d
 dx
                                                 
 (ϕ)sin(ϕ) = sin(ϕ)

                                                                   dx dϕ

                                                            = sin(ϕ)G (x)[−sin(ϕ)]
                                                                                    2
                                                                  2

                                                            =−sin (ϕ)G (x) =−[1 − cos (x)]G (x)

                                                                    2

                                                            =−(1 − x )G (x).
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                                   October 14, 2010  15:27  THM/NEIL   Page-656        27410_18_ch18_p641-666
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