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6.7 The Korteweg–de Vries Equation  263

                                    u rs =  (τ rs ) y  ,
                                           τ rs
                                    v rs =  (τ rs ) x  ,
                                           τ rs
          for all values of n and all differentiable functions f rs (x, y).
          Proof.

                                   1
                          (q rs ) y =     (τ r+1,s ) y  (τ rs ) y
                                  τ  2
                                   rs  τ r+1,s   τ rs

                                =  τ r+1,s (τ r+1,s ) y  −  (τ rs ) y
                                                    τ rs
                                   τ rs
                                          τ r+1,s
                                = q rs (u r+1,s − u rs ),
          which proves (a).
                                           G 11
                                   (u rs ) x =  ,
                                            τ 2
                                             rs
                                               G 13
                              v r+1,s − v rs = −  τ r+1,s τ rs  ,
                                              G 31
                              u rs − u r,s−1 =     ,
                                           τ rs τ r,s−1
                                               G 33
                               q rs − q r,s−1 = −    .
                                             τ rs τ r,s−1
          Hence, referring to (6.2.13),

                                                     G 11 G 33
                                                   =
                            (q rs − q r,s−1 )(u rs ) x
                        q rs (u rs − u r,s−1 )(v r+1,s − v rs )  G 31 G 13
                                                   =1,
          which proves (b).


          6.7 The Korteweg–de Vries Equation


          6.7.1  Introduction
          The KdV equation is
                                u t +6uu x + u xxx =0.               (6.7.1)
          The substitution u =2v x transforms it into
                                       2
                                 v t +6v + v xxx =0.                 (6.7.2)
                                       x
          Theorem 6.13. The KdV equation in the form (6.7.2) is satisfied by the
          function
                                   v = D x (log A),
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