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book   Mobk070    March 22, 2007  11:7








                                        SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS          71
                        The second solution of Legendre’s equation can be found by the method of variation of
                   parameters. The result is


                                                                   dζ

                                              Q n (x) = P n (x)                                (4.141)
                                                                         2
                                                                2
                                                              P (ζ)(1 − ζ )
                                                               n
                   It can be shown that this generally takes on a logarithmic form involving ln [(x + 1)/(x − 1)]
                   which goes to infinity at x = 1. In fact it can be shown that the first two of these
                   functions are

                                             1   1 + x              x   1 + x
                                       Q 0 =   ln        and  Q 1 =   ln      − 1              (4.142)
                                             2   1 − x              2   1 − x

                   Thus the complete solution of the Legendre equation is

                                                  u = AP n (x) + BQ n (x)                      (4.143)


                   where P n (x)and Q n (x) are Legendre polynomials of the first and second kind. If we require
                   the solution to be finite at x = 1, B must be zero.
                        Referring back to Eqs. (3.46) through (3.53) in Chapter 3, we note that the eigenvalues
                   λ = n(n + 1) and the eigenfunctions are P n (x)and Q n (x). We further note from (3.46) and
                   (3.47) that the weight function is one and that the orthogonality condition is

                                               1
                                                                   2

                                                 P n (x)P m (x)dx =    δ mn                    (4.144)
                                                                 2n + 1
                                              −1
                   where δ mn is Kronecker’s delta, 1 when n = m and 0 otherwise.

                   Example 4.13. Steady heat conduction in a sphere

                   Consider heat transfer in a solid sphere whose surface temperature is a function of θ, the angle
                   measured downward from the z-axis (see Fig. 1.3 Chapter 1). The problem is steady and there
                   is no heat source.

                                              ∂ 2        1   ∂       ∂u
                                            r    (ru) +          sin θ    = 0
                                              ∂r  2     sin θ ∂θ     ∂θ
                                            u(r = 1) = f (θ)                                   (4.145)

                                            u is bounded
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