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P. 770

750    CHAPTER 22  Singularities and the Residue Theorem

                                 Now use the facts that

                                                                 I (z) = iJ (iz)

                                                                  0      0
                                 and

                                                             J (z) =−J 1 (z) = J 1 (−z)
                                                              0
                                 to obtain
                                                            Res(g(z)/h(z), j n )

                                                              −2R J 0 ( j n r/R)  2  2
                                                            =              e − j n t/R  .
                                                                     J 1 ( j n )
                                                                j n
                                 The solution is therefore

                                                                   ∞
                                                                                     2
                                                                      −2R J 0 ( j n r/R)  − j n t/R  2
                                                   u(r,t) = T 0 1 − 2             e       .
                                                                       j n  J 1 ( j n )
                                                                   n=1
                        SECTION 22.4        PROBLEMS


                                                                         1
                     In each of Problems 1 through 10, use Theorem 22.7 to  5.
                     find the inverse Laplace transform of the function.  (z + 5) 3
                                                                         1
                                                                    6.
                                                                       z + 8
                                                                        3
                          z                                              1
                      1.                                            7.
                                                                        4
                         2
                        z + 9                                          z + 1
                           1                                              1
                      2.                                            8.  z
                        (z + 3) 2                                      e (z − 1)
                             1                                           z  2
                      3.                                            9.
                        (z − 2) (z + 4)                                (z − 2) 3
                             2
                             1                                            z + 3
                      4.                                           10.
                                                                        3
                          2
                        (z + 9)(z − 2) 2                               (z − 1)(z + 2)


























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                                   October 14, 2010  15:37  THM/NEIL   Page-750        27410_22_ch22_p729-750
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