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,
                                                                                          Unit 1: Bessel s Functions




                                                                                                Notes
                                                  2r n
                                     ( 1) r    x
                            n
                       = ( 1)
                                 (n r  1) (r  1) 2
                             r  0
                            n
          or       J (x) = ( 1)  J (x)                                            ...(26)
                    n        n
          Thus J (x) is not independent of J (x)
                n                    n
                                   ,
          1.3.2 Solution of Bessel s Differential Equation when n is a Non-negative
                Integer
          We had seen that when n is not an integer there are two independent solutions i.e. J (x) and J (x).
                                                                           n       n
          When n is a non-negative integer
                            n
                   J (x) = ( 1)  J (x)                                            ...(26)
                    n        n
          And so it is dependent on J (x). To find a second solution we introduce Neumann Function
                                n
                         J  ( )cos v J  ( )
                                       x
                           x
                   Y (x) =  v        v                                            ...(29)
                    v         sin v
                                                                        ,
          If v is not an integer, then Y (x) and J (x) form a general solution of the Bessel s equation. If v is
                                v       v
          a non-negative integer, then from equation (26), equation (29) becomes an indeterminate form.
          To calculate the limit of (29) for v    n, differentiate both the numerator and denominator with
          respect to v. Then setting v   n, we have

                                sin  J  ( ) cos  J  ( ) J v  ( )
                                                        x
                                                  x
                                       x
                          lim
             x
           Y n ( )  limY  ( ) =   n       cos
                      x
                   n
                                               x
                               x
                           1 J  ( )   ( 1) n  J  ( )
                         =                                                       ...(29a)
                                   n               n
          Now from equation (21)
                                ( 1)s   x  2s
                     J (x) =
                              (  s  1) s  2
                           s  0
                    J  ( )   ( 1)s x  2s  x 2  log  x 2  (  s  1)  x
                      x
                         =                                2
                               s  2     (  s  1)   (u s  1)  2
                           s  0
                                  x
                                s
                             ( 1) ( /2)  2     x
                         =                 log      (  s  1) ]
                               s      s  1    2
                           s  0
          where
                            (  s  1)
                 (  s  1) =                                                       ...(30)
                            (  s  1)

                      x
                    J  ( )   ( 1) s x 2  n  2s  x
          thus  lim      =             log      (n s  1)                          ...(31)
                  n          s  n s  1     2
                           s  0

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