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508    CHAPTER 15  Special Functions and Eigenfunction Expansions

                                 Case 2: λ< 0
                                          2
                                                        kx
                                 Say λ =−k .Now y(x) = c 1 e + c 2 e −kx . The condition y(L) = y(−L) gives us
                                                                    kL
                                                                          kL
                                                          c 1 e  −kL  + c 2 e = c 1 e + c 2 e −kL .
                                 Write this as
                                                                                 kL
                                                                   kL
                                                          c 1 (e −kL  − e ) = c 2 (e  −kL  − e ).
                                                                kx


                                 This implies that c 1 = c 2 ,so y(x) = c 1 (e + e −kx ).Now y (−L) = y (L) gives us
                                                                   kL
                                                                            kL
                                                         c 1 k(e −kL  − e ) = c 1 k(e − e −kL ).
                                 This implies that c 1 =−c 1 , hence c 1 =0, so c 2 =0 also. This problem has only the trivial solution,
                                 so there is no negative eigenvalue.
                                 Case 3: λ> 0

                                        2
                                                                  2
                                 Say λ = k . The general solution of y + k y = 0is

                                                           y(x) = c 1 cos(kx) + c 2 sin(kx).
                                 Now
                                            y(−L) = c 1 cos(kL) − c 2 sin(kL) = y(L) = c 1 cos(kL) + c 2 sin(kL).
                                 This implies that c 2 sin(kL) = 0. Next,



                                         y (−L) = kc 1 sin(kL) + kc 2 cos(kL) = y (L) =−kc 1 sin(kL) + kc 2 cos(kL),
                                 implying that kc 1 sin(kL) = 0. If sin(kL)  = 0, then c 1 = c 2 = 0 and we have only the trivial
                                 solution. For a nontrivial solution, we must have sin(kL) = 0, so at least one of the constants c 1
                                 and c 2 can be chosen nonzero. But sin(kL) = 0 is satisfied if kL = nπ, with n a positive integer
                                 (positive because we chose k > 0). Since λ = k , the eigenvalues of this problem, indexed by n,
                                                                      2
                                 are
                                                                 nπ
                                                                     2
                                                           λ n =      for n = 1,2,···
                                                                 L
                                 and corresponding eigenfunctions are
                                                                    nπx          nπx

                                                       y n (x) = c 1 cos  + c 2 sin   ,
                                                                     L            L
                                 with c 1 and c 2 any constants, not both zero.
                                    By choosing n = 0 and c 2 = 0but c 1  = 0, we obtain a constant eigenfunction corresponding
                                 to the eigenvalue 0. This consolidates cases 1 and 3.

                                    Bessel’s equation will provide an example of a singular Sturm-Liouville problem (here we
                                 will have r(0) = 0).
                                    We may also have a Sturm-Liouville problem in which r(a) = r(b) = 0, but there are no
                                 boundary conditions specified. In this event we impose the condition that eigenfunctions must
                                 be bounded on [a,b]. The Legendre differential equation will provide an example of this type of
                                 problem.
                                    A Fourier sine series on [0, L] is an expansion in the eigenfunctions of Example 15.1. A
                                 Fourier series on [−L, L] is an expansion in the eigenfunctions of Example 15.2. This raises an
                                 intriguing question. If a Sturm-Liouville problem on [a,b] has eigenfunctions ϕ 1 (x),ϕ 2 (x),···,




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                                   October 14, 2010  15:20  THM/NEIL   Page-508        27410_15_ch15_p505-562
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