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Unit 29: Fredholm Equations Solution by the Method of Successive Approximation




          This is called Neumann Series. Substituting (2) into (1) we obtain                    Notes
                                            1
                           x
                  f  ( )  1 ( )  2  2 ( ) ....  K ( , ) f  ( )  1 ( )  2  2 ( ) .... du  f  ( ) ...(3)
                   x
                                   x
                                              x
                                                                    x
                                                                               x
                                                            u
                                                u
                                                     u
                                           0
          Comparing the powers of   on both sides we have
                         1
                            x
                    ( )   K ( , ) ( )du  0
                    x
                               f
                              u
                                 u
                   1
                         0
                             x
                                 f
                                  u
                               u
                      x
                     ( )   K ( , ) ( )du
                    1
                           1               1      1
                      x
                     ( )    K ( , )  ( )du  K ( , )  K ( , u  ) (u  )du
                                    u
                                                     u
                              x
                                               u
                                                         f
                                              x
                                u
                    2              1                    1   1  1
                           0               0      0
                           1
                            K 2 ( , y 1 ) (u 1 )du
                               x
                                    f
                           0
                           1               1
                                    u
                                                   u
                                                u
                                u
                                              x
                              x
                            K ( , )  ( )du  K  ( , ) ( )du
                                                  f
                       3           2         3
                           0              0
                  
                           1                1
                                                  u
                            K ( , )  ( )du   K  ( , )du    (for n  1, 2, ...)
                              x
                                                x
                                u
                                      u
                      n            n  1       n
                           0                0
          In the above we have
                    x
                      u
                                      u
                  K  2 ( , )  K ( , u 1 ) (u 1 , )du 1
                             x
                                 K
                    x
                  K  ( , )  K ( , u K  (u  , )du  .                                ...(4)
                             x
                                       u
                      u
                                )
                   3            1  2  1   1
                  ........................................................
          and so on.
          More generally
                                         u
                              x
                    x
                  K  ( , )  K  ( , u  )K  (u  , )du  [n = 2, 3, 4, ...; r = 1, 2,..., n   1; K  = K ...(5)
                      u
                                   n
                                       1
                                     r
                   n
                            r
                                1
                                            1
                                                                             1
          Thus the series for the resolvent Kernel H(x, u,  ) is given by
                                                               u
                                                             x
                                         u
                                                 x
                                                   u
                                       x
                              x
                                u
                   H ( , , ) K ( , )  K  2 ( , )  2 K  3 ( , ) ...  n K  n ( , )   ...(6)
                     x
                       u
          The solution then is given by
                                       f
                                  x
                         x
                                         u
                  y ( )  f ( )  H ( , u , ) ( )du                                  ...(7)
                   x
          The main difference from the Volterra case is that the series for the resolvent Kernel (6) now
          converges only for sufficiently small values of | |. In other words, although H(x, u,  ) is still
          analytic function of   it is no longer an entire function of  .
          29.2 Lower Bound for the Radius of Convergence
          We shall now determine a lower bound for the radius of convergence of the power series (6).
          We observe that if we preserve the basic hypothesis i.e. that the Kernel K(x, y) is an L  Kernel,
                                                                               2
                                                  u
                            x
                              u
                                         x
          i.e.    K  2   K  2 ( , )dxdu  A 2 ( )dx  B 2 ( )du  N  2                ...(8)
          where
                                   1 2                1 2
                         1                   1
                                      B
                                       u
                    x
                             x
                                                x
                                                  u
                  A ( )   K  2 ( , )du  , ( )  K 2 ( , )dx                         ...(9)
                              u
                         0                  0
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