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Differential and Integral Equation




                    Notes          we have successively
                                                                    2
                                               u
                                                      x
                                          K 2 2 ( , )  K ( , u 1 ) (u 1 , )du 1  A  2 ( ) B 2 ( )
                                             x
                                                                          x
                                                                              u
                                                          K
                                                               u
                                          K 3 2 ( , )  K 2 ( , u 1 )du 1  K 2 2 (u 1 , )du 1  A 2 ( ) B 2 ( ) A 2 (u 1 )du 1  A 2 ( ) B 2 ( )N 2
                                                                                 u
                                                                             x
                                                                    u
                                                      x
                                                                                                 x
                                                                                                     u
                                               u
                                             x
                                                        z
                                                               z
                                             x
                                               u
                                                      x
                                                                              u
                                                                         x
                                                                                      z
                                                                                                   u
                                                                                              x
                                                                 u
                                          K  2 ( , )  K  2 ( , )dz K 2 ( , )dz  A  2 ( ) B 2 ( ) N  2  A  2 ( )dz  A 2 ( ) B 2 ( )N  4
                                           4                  3
                                   and hence in general
                                                         B
                                                          x
                                                       x
                                                 u
                                               x
                                          K   ( , )  A ( ) ( )N  n                           (n = 0, 1, 2, ....) ...(10)
                                            n  2
                                   If we neglect the first term of (6), this process that (6) has the majorant
                                                             n
                                            x
                                                x
                                                           N
                                          A ( ) ( )| |  (| | ) ;
                                              B
                                                     n  0
                                   This is a geometric series with the common ratio | |N, hence it converges for | |N < 1, i.e. for
                                                 1
                                              K                                                           ...(11)
                                   We thus see that under the condition (11), the partial sums of (6) have a majorant of the type
                                                                    C A(x) B(x)
                                   where C is a constant i.e., a majorant which is L  function of both x and u. In other words (6) is an
                                                                        2
                                   almost uniformly convergent series,  hence a series which can be integrated term-by-term  in
                                   either  x or  u (by Lebesgue fundamental theorem). Now the resolvent Kernel is an  analytic
                                   function whose singular points are outside or on the boundary of the circle (11).
                                   Since term-by-term integration is permitted, we see that by using (5) under condition (11) we
                                   have
                                            1
                                             K ( , u 1 ) (u 1 , , )du 1  H ( , u 1 , ) (u 1 , )du 1
                                                                     x
                                                                                u
                                                   H
                                               x
                                                        u
                                                                            K
                                            0
                                                                K 2 ( , )  K 3 ( , ) ...
                                                                            x
                                                                              u
                                                                   x
                                                                     u
                                                                  1
                                                                               x
                                                                   H
                                                                  [ ( , , )  K ( , )]
                                                                     x
                                                                                u
                                                                       u
                                   that is,
                                                                        u
                                                    x
                                              u
                                            x
                                                     u
                                          K ( , )  H ( , , )  K ( , u H (u  1 , , )du  1
                                                               x
                                                                   )
                                                                  1
                                                                                                          ...(12)
                                                             H ( , u  , ) (u  , )du
                                                                          u
                                                                      K
                                                               x
                                                                  1      1    1
                                   Now considering that all the terms of this double equality are analytic functions of  , we can
                                   thus assert that the basic equation (12) for the resolvent Kernel are valid not only in the circle
                                   (11), but in the whole domain of existence of the resolvent Kernel H in the   plane. If now f(x)
                                   belongs to the class L , then the given equation (1) has at least one solution of the same class L ,
                                                    2                                                         2
                                   this solution is
                                                                u
                                                           u
                                                 x
                                                         x
                                          y ( )  f  ( )  H ( , , ) ( )du                                  ...(13)
                                            x
                                                               f
                                   in the domain of existence H, of H.
                                   Moreover, it is easy to see that the solution (13) is the unique L -solution of our equation, not
                                                                                      2
                                   only inside the circle   K  1   but also in the whole domain of existence H.
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