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6.5 The Toda Equations  253

          are proved in Theorem 4.30 in Section 4.8.5 on Turanians.
            Let the elements in both A n and B n be defined as
                           φ m (x)= f (m) (x),  f(x) arbitrary,

          so that
                                     φ    = φ m+1                    (6.5.4)
                                      m
          and both A n and B n are Wronskians (Section 4.7) whose derivatives are
          given by
                                           (n+1)

                                   A = −A       ,
                                     n     n+1,n
                                           (n+1)

                                   B = −B       .                    (6.5.5)
                                     n     n+1,n
          Theorem 6.1. The equation
                                         u n u n+1

                                    u =
                                     n
                                          u n−1
          is satisfied by the function defined separately for odd and even values of n
          as follows:
                                   u 2n−1 =  A n  ,
                                           B n−1
                                     u 2n =  B n  .
          Proof.                           A n

                           B 2  u     = B n−1 A − A n B
                             n−1 2n−1         n       n−1
                                               (n+1)       (n)
                                      = −B n−1 A    + A n B n,n−1
                                               n+1,n

                     B 2   u 2n−1 u 2n  = A n−1 B n .
                      n−1
                             u 2n−2
          Hence, referring to (6.5.3),

               2
             B n−1  u 2n−1 u 2n  − u   2n−1  = A n−1 B n + B n−1 A (n+1)  − A n B (n)
                                                                  n,n−1
                     u 2n−2                            n+1,n
                                     =0,
          which proves the theorem when n is odd.
                                 2

                               A u    = A n B − B n A
                                 n 2n      n       n
                                             (n+1)      (n+1)
                                     = −A n B     + B n A n,n+1 ,
                                             n+1,n

                      A 2  u 2n u 2n+1  = A n+1 B n−1 .
                        n
                            u 2n−1
          Hence, referring to (6.5.2),

               A 2  u 2n u 2n+1  − u    = A n+1 B n−1 + A n B (n+1)  − B n A (n+1)
                 n             2n                    n+1,n      n,n+1
                     u 2n−1
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