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Unit 22: Solution of Laplace Differential Equation




                                                                                                Notes
                       2
                          2
                                               2
                       r d R  r dR  2 2  =   1 d   µ 2                            ...(12)
                                     r
                       R dr  2  R dr          d  2
          which gives
                    2
                  2 d R  dR    2 2  2
                 r   2  r    (  r  µ  )R = 0                                      ...(13)
                   dr    dr
          and
                               d 2   2
                                    µ   = 0                                       ...(14)
                               d  2

                                               x
          In equation (13) if we use the substitution  r  ,  it reduces to

                     2
                    d R  1 dR  1  µ 2  R = 0                                      ...(15)
                     dx 2  x dx   x 2

          Equation (15) is Bessel s differential equation and so the solution is given by
                                      R = A J  µ ( ) B J –µ ( )
                                               x
                                                       x
          or                          R = A J  µ ( r ) B J –µ ( r )               ...(16)

          where µ is not an integer and

                                      R = A J µ ( r ) B Yy  ( r )                 ...(17)
                                                     1
                                            1
          when µ is an integer. The solutions of equations (10), (14) are given by
                                      Z = A e  .z  B e  z                         ...(18)
                                            2
                                                   2
          and                           = A 3  cos(µ  ) B 3  sin (µ  )            ...(19)
          Hence the total solution is

            V = R    Z   A J  µ ( r ) B J  –µ ( r )  A e  z  B e  z  A 3  cos(µ  ) B 3  sin(µ  )  ...(20)
                                                 2
                                            2
          where µ is a fraction and  = 1, 2, 3... and

            V = R Z      A J  µ ( r ) BY  ( r )  A 3  cos(µ  ) B 3  sin( µ  ) A e  z  B e  z  ...(21)
                          1
                                                                   2
                                                              2
          When µ is an integer and   = 1, 2,...
          The solutions (20) and (21) depend upon the parameters µ,  . If we see a solution that is finite at
          r = 0 and also be single valued in   then µ be a positive integer and taking all values from 0 to  .
          Thus for a fixed  ,


                           )
            V =     A J  ( r A 3  cos µ  A 4 sin µ  A e  z  A e  z                ...(22)
                                                     2
                     1 µ
                                               2
                  µ  0
          Thus the above solution is known as cylindrical Harmonics and will be useful for certain physical
          problems.
          The solution (22) V for a single value of µ is called general cylindrical Harmonics.



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