Page 295 - Determinants and Their Applications in Mathematical Physics
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280   6. Applications of Determinants in Mathematical Physics

                           −H i+2,j + H i,j+2 .                     (6.8.20)
          From (6.8.15),
                                 v = h 00 − constant.

          The derivatives of v can now be found in terms of the h ij and H ij with the
          aid of (6.8.20):
                             2
                    v x = H 10 h ,
                             00
                                                3
                   v xx = H 20 + H 11 − 3h 00 H 10 +2h ,
                                                00
                            2
                  v xxx =12h H 10 − 3H 2
                            00        10  − 4h 00 H 20 − 3h 00 H 11 +3H 21
                                           4
                         +H 30 − 2h 10 h 01 − 6h ,
                                           00
                            ¯     ¯
                    v y = h 00 H 10 − H 20
                   v yy =2(h 10 h 20 + h 01 h 02 ) − (h 10 h 02 + h 01 h 20 )
                         −h 00 (h 2         2      2
                                                   00
                               10  − h 10 h 01 + h )+2h h 11
                                            01
                         −2h 00 H 21 + H 22 + h 00 H 30 − H 40 ,
                    v t =4(h 00 H 20 − h 10 h 01 − H 30 ).          (6.8.21)
          Hence,
                     2            2    2
               v t +6v + v xxx =3(h 10  + h 01  − h 00 H 11 + H 21 − H 30 ).  (6.8.22)
                     x
          The theorem appears after differentiating once again with respect to x.
          6.8.2  The Wronskian Solution

          The substitution
                                          2
                                   u =2D (log w)
                                          x
          into the KP equation yields

                                                        G
                                                     2
                       (u t +6uu x + u xxx ) x +3u yy =2D   ,       (6.8.23)
                                                        w
                                                     x   2
          where
                                                                    2
              G = ww xt − w x w t +3w 2  − 4w x w xxx + ww xxxx +3(ww yy − w ).
                                  xx                                y
          Hence, the KP equation will be satisfied if
                                       G =0.                        (6.8.24)
          The function G is identical in form with the function F in the first line
          of (6.7.60) in the section on the KdV equation, but the symbol y in this
          section and the symbol z in the KdV section have different origins. In this
          section, y is one of the three independent variables x, y, and t in the KP
          equation whereas x and t are the only independent variables in the KdV
          section and z is introduced to facilitate the analysis.
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