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748    CHAPTER 22  Singularities and the Residue Theorem



                         EXAMPLE 22.22
                                 Let
                                                                         1
                                                             F(z) =             .
                                                                   (z − 4)(z − 1) 2
                                                                     2
                                 Then F(z) has simple poles at ±2 and a double pole at 1. Compute
                                                                           e tz      1
                                                                                       2t
                                                         tz
                                                     Res(e F(z),2) = lim           = e ,
                                                                   z→2 (z + 2)(z − 1) 2  4
                                                                           e tz       1
                                                      tz
                                                  Res(e F(z),−2) = lim            =−    e  −2t ,
                                                                  z→−2 (z − 2)(z − 1) 2  36
                                 and
                                                                   d   2   −1 tz
                                                     tz
                                                Res(e F(z),1) = lim  ((z − 4) e )
                                                               z→1 dz
                                                                            −2 tz
                                                                       2
                                                                                      2
                                                                                    tz
                                                                                           −1
                                                             = lim(−2z(z − 4) e + te (z − 4) )
                                                               z→1
                                                                 1     2
                                                                         t
                                                                    t
                                                             =− te − e .
                                                                 3     9
                                 Then
                                                                1     1      1     2
                                                      −1          2t     −2t  − te − e .
                                                                                t
                                                                                     t
                                                     L [F](t) = e −     e
                                                                4    36      3     9
                                 22.4.1 Diffusion in a Cylinder
                                 We will find the temperature distribution function for a homogeneous, solid cylinder of radius R
                                 centered along the z-axis. This problem was solved in Section 17.5 using separation of variables.
                                 We will now use the Laplace transform and Theorem 22.7 to obtain the temperature distribution
                                 function. In the course of this, we will use properties of the modified Bessel function I 0 (x) and
                                 the Bessel functions J 0 (x) and J 1 (x) of the first kind of orders zero and one, respectively. These
                                 functions are developed in Section 15.3.
                                    We will assume angular independence and use the heat equation in cylindrical coordinates.
                                 The boundary value problem is
                                                         ∂u   ∂ u   1 ∂u
                                                               2
                                                            =    +      for 0 ≤r ≤ R,t > 0
                                                         ∂t   ∂r 2  r ∂r
                                                      u(r,0) = 0,u(R,t) = T 0 .
                                 Apply the Laplace transform with respect to t to this problem to obtain
                                                            2
                                                           ∂ U    1 ∂U
                                                               +      − sU(r,s) = 0.
                                                            ∂r  2  r ∂r
                                 This is a modified Bessel equation of order zero. A solution that is bounded at r = 0, which is
                                 the center of the cylinder, is given by
                                                                          √
                                                               U(r,s) = cI 0 ( sr),

                                 where I 0 (z) = J 0 (iz). Transform the condition u(R,t) = T 0 to obtain U(R,s) = T 0 /s. Then
                                                                        √      T 0
                                                            U(R,s) = cI 0 ( sR) =  .
                                                                                s



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                                   October 14, 2010  15:37  THM/NEIL   Page-748        27410_22_ch22_p729-750
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