Page 366 - Schaum's Outline of Theory and Problems of Advanced Calculus
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CHAP. 13]                         FOURIER SERIES                                357


                           Since B ¼ 0 makes the solution (8)identically zero, we avoid this choice and instead take
                                                                               m
                                                 sin 2  ¼ 0;  i.e.,  2  ¼ m   or    ¼                ð10Þ
                                                                               2
                           where m ¼ 0;  1;  2; .. . .
                              Substitution in (8) now shows that a solution satisfying the first two boundary conditions is
                                                                  2 2    m x
                                                                 3m   t=4
                                                      Uðx; tÞ¼ B m e  sin
                                                                          2                          ð11Þ
                           where we have replaced B by B m ,indicating that different constants can be used for different values of m.
                              If we now attempt to satisfy the last boundary condition Uðx; 0Þ¼ x; 0 < x < 2, we find it to be
                           impossible using (11).  However, upon recognizing the fact that sums of solutions having the form (11)
                           are also solutions (called the principle of superposition), we are led to the possible solution
                                                             1
                                                                    2 2
                                                            X     3m   t=4  m x
                                                               B m e   sin
                                                                           2
                                                     Uðx; tÞ¼                                        ð12Þ
                                                            m¼1
                              From the condition Uðx; 0Þ¼ x; 0 < x < 2, we see, on placing t ¼ 0, that (12)becomes
                                                        1
                                                        X       m x
                                                           B m sin    0 < x < 2
                                                                 2
                                                     x ¼                                             ð13Þ
                                                        m¼1
                           This, however, is equivalent to the problem of expanding the function f ðxÞ¼ x for 0 < x < 2intoa sine
                                                                                            4
                           series.  The solution to this is given in Problem 13.12(a), from which we see that B m ¼  cos m  so that
                                                                                          m
                           (12)becomes
                                                             4                 m x

                                                        1
                                                                        2 2
                                                       X               3m   t=4
                                                               cos m  e     sin
                                                            m                   2
                                                Uðx; tÞ¼                                             ð14Þ
                                                       m¼1
                           which is a formal solution.To check that (14)isactually a solution, we must show that it satisfies the partial
                           differential equation and the boundary conditions.  The proof consists in justification of term by term
                           differentiation and use of limiting procedures for infinite series and may be accomplished by methods of
                           Chapter 11.
                              The boundary value problem considered here has an interpretation in the theory of heat conduction.
                                           2
                                     @U   @ U
                           The equation  ¼ k  is the equation for heat conduction in a thin rod or wire located on the x-axis
                                     @t    @x 2
                           between x ¼ 0 and x ¼ L if the surface of the wire is insulated so that heat cannot enter or escape. Uðx; tÞ is
                           the temperature at any place x in the rod at time t.  The constant k ¼ K=s  (where K is the thermal
                           conductivity, s is the specific heat, and   is the density of the conducting material) is called the diffusivity.
                           The boundary conditions Uð0; tÞ¼ 0 and UðL; tÞ¼ 0indicate that the end temperatures of the rod are kept
                           at zero units for all time t > 0, while Uðx; 0Þ indicates the initial temperature at any point x of the rod. In
                           this problem the length of the rod is L ¼ 2 units, while the diffusivity is k ¼ 3 units.
                     ORTHOGONAL FUNCTIONS
                     13.25. (a) Show that the set of functions
                                                x     x     2 x    2 x    3 x     3 x
                                          1; sin  ; cos  ; sin  ; cos  ; sin  ; cos  ; ...
                                               L      L     L       L      L       L
                           forms an orthogonal set in the interval ð L; LÞ.
                           (b) Determine the corresponding normalizing constants for the set in (a)so that the set is
                           orthonormal in ð L; LÞ.
                           (a) This follows at once from the results of Problems 13.2 and 13.3.
                           (b)By Problem 13.3,
                                                 L   2 m x          L   2 m x
                                                ð                  ð
                                                   sin    dx ¼ L;     cos    dx ¼ L
                                                  L    L             L    L
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