Page 50 - Essentials of applied mathematics for scientists and engineers
P. 50

book   Mobk070    March 22, 2007  11:7








                     40  ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS























                       FIGURE 3.4: Eigenvalues of − tan(λ) = Hλ

                       Solutions exist only for discrete values λ n the eigenvalues. The corresponding solutions X n (x)
                       are the eigenfunctions.


                       3.3.1  Orthogonality of Eigenfunctions
                       Consider two solutions of (3.46) and (3.47), X n and X m corresponding to eigenvalues λ n and
                       λ m . The primes denote differentiation with respect to x.


                                                    (rX ) + qX m =−λ m pX m                         (3.48)
                                                       m

                                                     (rX ) + qX n =−λ n pX n                        (3.49)
                                                        n
                       Multiply the first by X n and the second by X m and subtract, obtaining the following:


                                              (rX n X − rX m X ) = (λ n − λ m )pX m X n             (3.50)
                                                    m         n
                       Integrating both sides

                                                                         b

                                                          b

                                          r(X X n − X X m ) = (λ n − λ m )  p(x)X n X m dx          (3.51)

                                                          a
                                                      n
                                              m
                                                                        a
                       Inserting the boundary conditions into the left-hand side of (3.51)
                          X (b)X n (b) − X (a)X n (a) − X (b)X m (b) + X (a)X m (a)




                                                                    n
                                         m
                            m
                                                       n
                                b 1             a 1             a 1             b 1
                            =−     X m (b)X n (b) +  X m (a)X n (a) −  X n (a)X m (a) +  X m (b)X n (b) = 0  (3.52)
                                b 2             a 2             a 2             b 2
   45   46   47   48   49   50   51   52   53   54   55