Page 503 - Electromagnetics
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Of course we could use exponentials or a combination of exponentials and trigonometric
                                                                                     2
                        functions instead. Rearranging (A.119) and multiplying through by ρ , we obtain
                                                                           2
                                                                              2
                                                2
                                             1 d #       2  2     2  ρ dP  ρ d P
                                           −        = k − k z  ρ +      +       .
                                             # dφ 2                P dρ   P dρ 2
                        The left and right sides depend only on φ and ρ, respectively; both must equal some
                                 2
                        constant k :
                                 φ
                                                             2
                                                          1 d #    2
                                                        −       = k ,                         (A.121)
                                                                   φ
                                                          # dφ  2
                                                                      2
                                                                    2
                                                           ρ dP    ρ d P
                                                 2  2    2                   2
                                               k − k z  ρ +     +         = k .               (A.122)
                                                                             φ
                                                           P dρ    P dρ 2
                        The variables ρ and φ are thus separated, and harmonic equation (A.121) has solutions

                                                    A φ sin k φ φ + B φ cos k φ φ,  k φ  = 0,
                                           #(φ) =                                             (A.123)
                                                    a φ φ + b φ ,         k φ = 0.
                        Equation (A.122) is a bit more involved. In rearranged form it is
                                                                    2
                                                 2
                                                d P   1 dP      2  k φ
                                                    +      + k −       P = 0                  (A.124)
                                                                c
                                                dρ 2  ρ dρ         ρ 2
                        where
                                                          2
                                                              2
                                                                  2
                                                         k = k − k .
                                                         c
                                                                  z
                        The solution depends on whether any of k z , k φ ,or k c are zero. If k c = k φ = 0, then
                                                        2
                                                       d P   1 dP
                                                           +      = 0
                                                       dρ 2  ρ dρ
                        so that
                                                      P(ρ) = a ρ ln ρ + b ρ .
                        If k c = 0 but k φ  = 0, we have
                                                     2
                                                    d P   1 dP   k 2 φ
                                                        +      −    P = 0
                                                    dρ 2  ρ dρ   ρ  2
                        so that
                                                    P(ρ) = a ρ ρ  −k φ  + b ρ ρ .             (A.125)
                                                                      k φ
                        This includes the case k = k z = 0 (Laplace’s equation). If k c  = 0 then (A.124) is Bessel’s
                                                                                                  (z)
                        differential equation. For noninteger k φ the two independent solutions are denoted J k φ
                               (z), where J ν (z) is the ordinary Bessel function of the first kind of order ν.For
                        and J −k φ
                        k φ an integer n, J n (z) and J −n (z) are not independent and a second independent solution
                        denoted N n (z) must be introduced. This is the ordinary Bessel function of the second
                        kind, order n. As it is also independent when the order is noninteger, J ν (z) and N ν (z)
                        are often chosen as solutions whether ν is integer or not. Linear combinations of these
                        independent solutions may be used to produce new independent solutions. The functions




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