Page 238 - Determinants and Their Applications in Mathematical Physics
P. 238

5.7 The One-Variable Hirota Operator  223

          Proof. First proof (Caudrey). The Hessenbergian satisfies the recurrence
          relation (Section 4.6)

                                      n
                                          n
                              E n+1 =        u r+1 E n−r .           (5.7.7)
                                          r
                                     r=0
          Let
                           H (f, g)
                             n
                      F n =        ,  f = f(x),g = g(x),F 0 =1.      (5.7.8)
                              fg
          The theorem will be proved by showing that F n satisfies the same
          recurrence relation as E n and has the same initial values.
            Let
                                       zH
                                        (f,g)
                                      e
                                    
                                        fg
                                    
                                           n  n
                                     ∞ !
                                           H (f,g)
                                          z
                                K =       n!  fg                     (5.7.9)
                                     n=0
                                           n
                                     ∞ !
                                    
                                          z F n
                                             .
                                    
                                           n!
                                      n=0
          Then,
                                        ∞
                                 ∂K        z n−1
                                     =         F n                  (5.7.10)
                                  ∂z       (n − 1)!
                                       n=1
                                        ∞
                                            n
                                           z F n+1
                                     =            .                 (5.7.11)
                                             n!
                                       n=0
          From the lemma and (5.7.6),
                        f(x + z)g(x − z)      1
                   K =                 = exp  2  {φ(x + z)+ φ(x − z)
                           f(x)g(x)

                        +ψ(x + z) − ψ(x − z) − 2φ(x)} .             (5.7.12)
          Differentiate with respect to z, refer to (5.7.11), note that
                            D z (φ(x − z)) = −D x (φ(x − z))
          etc., and apply the Taylor relations (5.7.2) from the previous section. The
          result is
             ∞
                z F n+1        1
                 n
                       = D    {φ(x + z) − φ(x − z)+ ψ(x + z)+ ψ(x − z)} K
                  n!         2
             n=0

                           ∞   2n+1  2n+2      ∞      2n+1
                                                   2n
                              z    D     (φ)      z D     (ψ)
                       =                    +                  K
                                 (2n + 1)!           (2n)!
                           n=0                n=0

                           ∞   2n+1         ∞
                                               2n
                              z    u 2n+2     z u 2n+1
                       =                 +              K
                               (2n + 1)!        (2n)!
                           n=0             n=0
   233   234   235   236   237   238   239   240   241   242   243