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book   Mobk070    March 22, 2007  11:7








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                                                 CHAPTER 8


                         Solutions with Laplace Transforms






                   In this chapter, we present detailed solutions of some boundary value problems using the Laplace
                   transform method. Problems in both mechanical vibrations and diffusion are presented along
                   with the details of the inversion method.


                   8.1    MECHANICAL VIBRATIONS
                   Example 8.1. Consider an elastic bar with one end of the bar fixed and a constant force F per
                   unit area at the other end acting parallel to the bar. The appropriate partial differential equation
                   and boundary and initial conditions for the displacement y(x, t) are as follows:

                                               y ττ = y ζζ , 0 <ζ < 1,  t > 0
                                               y(ζ, 0) = y t (ζ, 0) = 0
                                               y(0,τ) = 0
                                               y ς (1,τ) = F/E = g

                        We obtain the Laplace transform of the equation and boundary conditions as

                                                       2
                                                      s Y = Y ςς
                                                      Y (s, 0) = 0
                                                      Y ς (s, 1) = g/s

                        Solving the differential equation for Y (s , ζ),

                                              Y (s ) = (A sinh ςs + B cosh ς s )

                        Applying the boundary conditions we find that B = 0and
                                                     g
                                                       = As cosh s
                                                     s
                                                             g
                                                     A =
                                                           2
                                                          s cosh s
                                                            g sinh ς s
                                                     Y (s ) =
                                                             2
                                                            s cosh s
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