Page 295 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION
P. 295

Differential and Integral Equation




                    Notes         i.e. on expansion
                                              F F 1  F F 1   F  F 1  F  F 1  F  F 1  F  F 1
                                             x  p    p  x   x  p    p  x    x  p   p  x   = 0             ...(12)
                                              1  1    1  1   2  2    2  2    3  3   3  3
                                  The equation (12) is generally written as  F ,F 1  0.

                                  Similarly
                                                                 , F F  0 and  F  ,F  0.
                                                                  2         1  2
                                  But these are linear equations having more than two independent variables. Here we have the
                                  following rule.

                                  Try to find two independent integrals, F  = a   and F  = a , of the subsidiary equations
                                                                   1  1     2   2
                                                dx   dp    dx    dp    x      dp
                                                  1    1     2    2     3  =    3                         ...(13)
                                                  F    F     F    F     F       F
                                                 p 1  x 1    p 2  x 2   p 3    x 3
                                  If F , F  satisfy the conditions
                                     1  2
                                                                       F  F   F  F
                                                           F 1  ,F 2    1  2   1  2   0,
                                                                 r  1,2,3  x r  p r  p r  x r
                                  and if the p’s can be found as functions of the x’s from
                                                                F = F   a  = F   a  = 0,
                                                                    1  1  2   2
                                  then integrate the equation formed by substituting these functions in
                                                               dz = p dx  + p dx  + p dx .
                                                                   1  1  2  2  3  3
                                  Examples of Jacobi Method

                                  1.   Solve
                                                                             2
                                                               2p x x  3p x 2  p p  0
                                                                        2 3
                                                                 1 1 3
                                                                             2 3
                                  Solution:
                                                           2
                                  Let     F  2p x x  3p x  2  p p  0                                       ...(1)
                                               1 1 3  2 3  2 3
                                  The subsidiary equations are

                                           dx 1  dp 1  dx 2  dp 2  dx  3  dp 3
                                             F    F     F    F     F    F
                                                                                                             (2)
                                            p 1  x 1    p 2  x 2   p 3  x 3
                                  Now

                                                     F          F          F     2        F
                                                         2x x  ,   2p x  ,     3x 3  2p p  ,  0,
                                                                     1 3
                                                           1 3
                                                                                     2 3
                                                     p 1       x 1        p 2            x 2
                                                               F    2  F
                                                                   p 2  ,  2p x  6p x
                                                                                  2 3
                                                                            1 1
                                                              p 3      x 3
          288                               LOVELY PROFESSIONAL UNIVERSITY
   290   291   292   293   294   295   296   297   298   299   300