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Unit 28: Integral Equations




               reduced to an algebraic system of linear equations. If the integral of the given integral  Notes
               equation is replaced by a suitable sum then instead of dividing the basic integral into sub-
               intervals of the same size, it may be useful to divide it according to the zeros of a certain
               polynomial of Legendre. This method is developed in most books on numerical methods.

          28.4 Description of Some Methods used in the solution of Fredholm

               Integral Equation

          In the next few units we are interested in studying the Fredholm integral equations. In the unit
          29, we study the Fredholm equations by the method of successive approximation. In this iteration
          method either the unknown function or the Kernel is iterated into a series known as Neumann’s
          series. The convergence of the series depends upon the iterative parameter and the nature of the
          Kernel as well as the function in the domain a  x  b, a  t  b when the Kernel K(x, t) and the
          function f(x) are square integrable. For this purpose the function as well as Kernel has continuous
          derivations. Then in the unit 31 we will study the solution of Fredholm equations with a special
          type of Kernels  known as Poincere Goursat Kernels (P.G.).  In the  light of P.G. Kernels the
          existence and uniqueness of the solution of Fredholm equations of both kinds. In the unit 32 the
          final unit the famous Fredholm theorem on the existence and uniqueness of the solutions is
          described along with the conditions put on the functions.


                 Example: Express the differential equation
                                 2
                                d y  9y  = 18x    y (0)  y   0                     ...(1)
                                dx 2                     2

          as an integral equation.
          Solution: Integrating (1) from 0 to x,

                        x  2    x          x
                         d y dx  9 y ( )du =  18xdx
                                   u
                         dx 2
                        0       0          0
                                x
                     dy  dy
                                   u
          or                   9 y ( )du = 9x 2
                     dx  dx  x  0
                                0
                                x
                                   u
          or         y  ( ) y  (0) 9 y ( )du = 9x 2
                       x
                                0
          Again integrating
                             x  t            3
                   x                       9x
                                   u
                  x
                y ( )  y  (0)x  9 dt y ( )du =
                   0                        3
                             0  0
          or
                             x      x
            y ( ) y (0) y  (0)x  9 y ( )du dt = 3x 3
              x
                                u
                             0      0






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