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264   6. Applications of Determinants in Mathematical Physics

          where

                               A = |a rs | n ,
                                            2
                              a rs = δ rs e r +  = a sr ,
                                          b r + b s
                                              3
                              e r = exp(−b r x + b t + ε r ).
                                              r
          The ε r are arbitrary constants and the b r are constants such that the b r +
          b s  =0 but are otherwise arbitrary.
            Two independent proofs of this theorem are given in Sections 6.7.2 and
          6.7.3. The method of Section 6.7.2 applies nonlinear differential recurrence
          relations in a function of the cofactors of A. The method of Section 6.7.3
          involves partial derivatives with respect to the exponential functions which
          appear in the elements of A.
            It is shown in Section 6.7.4 that A is a simple multiple of a Wronskian and
          Section 6.7.5 consists of an independent proof of the Wronskian solution.



          6.7.2  The First Form of Solution

          First Proof of Theorem 6.1.3. The proof begins by extracting a wealth
          of information about the cofactors of A by applying the double-sum rela-
          tions (A)–(D) in Section 3.4 in different ways. Apply (A) and (B) with
          f interpreted first as f x and then as f t . Apply (C) and (D) first with

                                        3
                                                                         2
          f r = g r = b r , then with f r = g r = b . Later, apply (D) with f r = −g r = b .
                                        r                                r
            Appling (A) and (B),

                          v = D x (log A)= −     δ rs b r e r A rs
                                            r  s

                                       = −    b r e r A ,            (6.7.3)
                                                   rr
                                            r

                               D x (A )=    b r e r A A .            (6.7.4)
                                                 ir
                                                    rj
                                    ij
                                          r
          Applying (C) and (D) with f r = g r = b r ,



                               δ rs (b r + b s )e r +2 A rs  =2  b r ,
                         r  s                          r
          which simplifies to

                        b r e r A rr  +  A rs  =  b r ,              (6.7.5)
                      r           r   s        r
                                                    1
                        b r e r A A rj  +  A A rj  = (b i + b j )A .  (6.7.6)
                                                             ij
                             ir
                                            is
                                                    2
                      r              r  s
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