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Chapter 8





               Initial Boundary Value Problem for

               the Wave Equation







               8.1      Solved Problems

               Exercise 159. Consider a string of length L with a fixed left end a free right end.
               Initially the string is at rest with displacement f(x). Find the motion of the string
               by solving,


                                                   2
                                           u = c u ,        0 < x < L,       t > 0,
                                             tt
                                                      xx
                                                u(0, t) = 0,    u (L, t) = 0,
                                                                  x
                                             u(x, 0) = f(x),       u (x, 0) = 0,
                                                                    t
               with separation of variables.

               Solution: Use separation of variables to find eigen-solutions of the partial differential
               equation that satisfy the homogeneous boundary conditions. There will be two
               eigen-solutions for each eigenvalue. Expand u(x, t) in a series of the eigen-solutions.
               Use the two initial conditions to determine the constants.


                                                   2
                                           u = c u ,        0 < x < L,       t > 0,
                                                      xx
                                             tt
                                                u(0, t) = 0,    u (L, t) = 0,
                                                                  x
                                             u(x, 0) = f(x),       u (x, 0) = 0,
                                                                    t
               We substitute the separation of variables u(x, t) = X(x)T(t) into the partial dif-
               ferential equation.


                                                                 2
                                                    (XT) = c (XT)        xx
                                                           tt
                                                      T  00   X 00
                                                          =       = −λ   2
                                                      2
                                                     c T      X
               With the boundary conditions at x = 0, L, we have the ordinary differential equa-
               tions,

                                                         00
                                                                  2 2
                                                       T = −c λ T,

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