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6.7 The Korteweg–de Vries Equation  267
                                 1
                                                  b )A
                           =0 +       (b i+2 j  j+2 i  pr
                                          b − b
                                 2     p   r   p   r
                                   p,r
                              1
                           =   (φ i+2,j − φ i,j+2 ),
                              2
          which is identical with the right side of (a). This completes the proof of
          (a).
            Referring to (6.7.8) with r, s → p, q and i, j → r, s,

                 D x (φ ij )=   b b D x (A )
                                         rs
                                 i j
                                 r s
                           r  s


                        =       b b   1 (b r + b s )A rs  −  A A ps
                                                            rq
                                 i j
                                      2
                                 r s
                           r  s                      p  q
                          1
                        =         b b (b r + b s )A rs  −  b A rq  b A ps
                                                               j
                                                       i
                                   i j
                          2        r s                 r       s
                             r  s                  q,r      p,s
                          1
                        =   (φ i+1,j + φ i,j+1 ) − φ i0 φ j0 ,
                          2
          which proves (b). Part (c) is proved in a similar manner.
            Particular cases of (a)–(c) are
                                        1
                          φ 00 φ 11 − φ 2  = (φ 21 − φ 03 ),        (6.7.18)
                                   10   2
                                              2
                              D x (φ 00 )= φ 10 − φ ,
                                              00
                              D t (φ 00 )=2φ 00 φ 20 − φ 2 10  − φ 30 .  (6.7.19)
          The preparations for finding the derivatives of v are now complete. The
          formula for v given by (6.7.7) can be written
                                 v = φ 00 − constant.
          Differentiating with the aid of parts (b) and (c) of the lemma,
                                     2
                          v x = φ 10 − φ ,
                                     00
                                                       3
                               1
                         v xx = (φ 20 + φ 11 − 6φ 00 φ 10 +4φ ),
                               2                       00
                                                         2
                               1
                        v xxx = (φ 30 +3φ 21 − 8φ 00 φ 20 − 14φ ,
                               4                         10
                                    2
                                                       4
                               +48φ φ 10 − 6φ 00 φ 11 − 24φ )
                                    00
                                                       00
                           v t =2φ 00 φ 20 − φ 2 10  − φ 30 .       (6.7.20)
          Hence, referring to (6.7.18),
                                                              2
                          2

                  4(v t +6v + v xxx )=3 (φ 21 − φ 30 ) − 2(φ 00 φ 11 − φ )
                                                              10
                          x
                                   =0,
          which completes the verification of the first form of solution of the KdV
          equation by means of recurrence relations.
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