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A.12 B¨acklund Transformations  337
          A.12    B¨acklund Transformations


          It is shown in Section 6.2.8 on brief historical notes on the Einstein and
          Ernst equations that the equations
                              1          2          2
                              2 (ζ + + ζ − )∇ ζ ± =(∇ζ ± ) ,
          where
                                              2
                               ζ ± = φ ± ωψ  (ω = −1),             (A.12.1)
          are equivalent to the coupled equations
                                2
                                         2
                                                 2
                             φ∇ φ − (∇φ) +(∇ψ) =0,                 (A.12.2)
                                     2
                                  φ∇ ψ − 2∇φ ·  ψ = , 0            (A.12.3)
          which, in turn, are equivalent to the pair
                              1            2    2    2   2

                                        − φ − φ + ψ + ψ =0,        (A.12.4)
                              ρ            ρ    z    ρ   z
                     φ φ ρρ + φ ρ + φ zz
                                     ∂  	    
   ∂
                                               +     ρρ z  =0.     (A.12.5)
                                         ρρ ρ
                                     ∂ρ   φ 2    ∂z   φ 2
          Given one pair of solutions of (A.12.1), it is possible to construct other
          solutions by means of B¨acklund transformations.
          Transformation δ
          If ζ + and ζ − are solutions of (A.12.1) and


                                    ζ = aζ − − b,
                                     +

                                    ζ = aζ + + b,
                                     −
          where a, b are arbitrary constants, then ζ     and ζ     are also solutions of
                                               +      −
          (A.12.1). The proof is elementary.
          Transformation γ
          If ζ + and ζ − are solution of (A.12.1) and
                                          c

                                    ζ =     + d,
                                     +
                                         ζ +
                                          c

                                    ζ =     − d,
                                     −
                                         ζ −

          where c and d are arbitrary constants, then ζ and ζ are also solutions of

                                                 +
                                                       −
          (A.12.1).
          Proof.
                                    c(ζ + + ζ − )
                        1 (ζ + ζ )=           ,


                        2  +   −
                                      2ζ + ζ −
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