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568    CHAPTER 16  The Wave Equation

                                 16.2.1 Zero Initial Velocity
                                 An elastic string of length L with fixed ends is released from rest (zero initial velocity) from
                                 an initial position given as the graph of y = f (x). The initial-boundary value problem for the
                                 position function y(x,t) is
                                                            2
                                                                   2
                                                          ∂ y     ∂ y
                                                              = c 2  for 0 < x < L,t > 0,
                                                           ∂t 2   ∂x 2
                                                        y(0,t) = y(L,t) = 0for t ≥ 0,
                                                        y(x,0) = f (x) for 0 ≤ x ≤ L,
                                 and
                                                                  ∂y
                                                                    (x,0) = 0.
                                                                  ∂t
                                 The Fourier method (or method of separation of variables) is to attempt a solution of the form
                                 y(x,t) = X(x)T (t). Substitute this into the wave equation to get
                                                                        2


                                                                 XT = c X T,
                                 where X = dX/dx and T = dT/dt. Then


                                                                   X      T
                                                                     =     .
                                                                        2
                                                                   X    c T
                                 The left side depends only on x. We could fix x and then the right side, which depends only
                                 on t, would be constant for all t. But then the left side must equal the same constant for all x.
                                 Therefore, for some number λ, called the separation constant,
                                                                X      T
                                                                   =     =−λ.
                                                                      2
                                                                 X   c T
                                 Calling the constant −λ is common practice. Then
                                                                              2

                                                          X + λX = 0 and T + λc T = 0,

                                 two ordinary differential equations for X and T . Next use the boundary conditions. First,
                                                          y(0,t) = X(0)T (t) = 0for t ≥ 0
                                 implies that X(0) = 0. Similarly, y(x, L) = X(L)T (t) = 0 implies that X(L) = 0. We have
                                 obtained a Sturm-Liouville problem for X:

                                                          X + λX = 0; X(0) = X(L) = 0.
                                 In Example 15.1, we solved this problem for the values of λ (eigenvalues) and corresponding
                                 solutions for X (eigenfunctions):
                                                        2
                                                      n π 2               nπx
                                                  λ n =    and X n (x) = sin  for n = 1,2,··· .
                                                       L  2               L
                                                            2
                                                              2
                                                                 2
                                    Next focus on T . Since λ n = n π /L , the differential equation for T is
                                                                       2 2
                                                                     2
                                                                    n π c

                                                               T +       T = 0.
                                                                      L  2
                                 Because the string is released from rest,
                                                            ∂y

                                                               (x,0) = X(x)T (0) = 0,
                                                            ∂t
                                 so
                                                                   T (0) = 0.

                                 Solutions for T (t) subject to this condition are constant multiples of cos(nπct/L).
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                                   October 14, 2010  15:23  THM/NEIL   Page-568        27410_16_ch16_p563-610
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