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Unit 13: Orthogonality of Solutions




          Thus we have obtained infinity many orthogonal sets corresponding to each fixed value of  n.  Notes
          If a function is represented by generalized Fourier Bessel series


                                               f(x) =   C J   mn x  ,   for n fixed  ...(iii)
                                                         m n
                                                      m  1
                                                               b
                                                          1
          then                                 C   =            x f ( )J  x dx , m  1,2...
                                                                   x
                                                 m            2      n  mn
                                                      J     x
                                                       n  mn   a
          Since                                p(x) = x,    mn   mn
                                                                 a

                                                      a
                                                 2
          where                           J n  mn x  =  x J n 2  mn x dx          ...(iv)
                                                      0
                                                2
          To bind                        J  n  mn x  ,

          let us proceed as follows:
                                ),
          Multiplying (i) by  2x J  n ( x  we have

                                  n 2        1
                      1
               1
                               2
                  )
            2x J n ( x x J n ( x )  x  2  2x J n ( x ) ( x ) = 0
                                             J
                                             n
                                  x
                               2    2 2  2
                           1
                            x
          or             x J  n ( )  x  n  J n ( x )  = 0
          Integrating over the limits 0 to a, we have
                                      a
                                    2
                             x J  1 n ( x )  =  2 2  n 2  J 2 n ( x ) dx
                                                x
                                      0
          Integrating R.H.S. by parts, we have
                                      a                    a     a
                                    2
                                                    2
                             x J 1 n  ( x )  =  2 2  n J  2 n  ( x )  2  2  x J  2 n ( x )dx  ...(v)
                                               x
                                      0                    0
                                                                 0
          From the following recurrence formulas for J ( ), we have
                                               n
                                       d   n             n
                                                             µ
                                               µ
                                          µ  J n ( )  =  µ  J  ( )
                                      dµ                  n  1
                               n d         n  1          n
                                                µ
                                     µ
                                                             µ
          or                 µ    J n ( ) n µ  J n ( ) =  µ  J n  1 ( )
                                dµ
          Multiplying both sides by   n+1
                                     d
                                       J n ( ) n J n ( ) =  µ J  ( )
                                                            µ
                                    d                    n  1
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