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18.2 Dirichlet Problem for a Rectangle  643


                                                                   y



                                                                (0, K)  u(x, K) = f(x)  (L, K)


                                                                 u = 0               u = 0
                                                                           D

                                                                                             x
                                                                         u = 0     (L, 0)



                                                              FIGURE 18.1 D and the boundary data
                                                              for this problem.


                                        and

                                                                     Y − λY = 0;Y(0) = 0.
                                        The problem for X has eigenvalues and eigenfunctions
                                                                     2
                                                                    n π  2              nπx
                                                                λ n =    and X n (x) = sin
                                                                     L  2                L
                                        for n = 1,2,···.
                                           The problem for Y becomes
                                                                         2
                                                                        n π  2

                                                                   Y −      Y = 0;Y(0) = 0
                                                                         L 2
                                        with solutions that are constant multiples of
                                                                                 nπy

                                                                     Y n (y) = sinh   .
                                                                                  L
                                           For each positive integer n, we have a function
                                                                             nπx       nπy

                                                                u n (x, y) = sin  sinh
                                                                              L         L
                                        that is harmonic on D and satisfies the zero boundary conditions on the lower side and vertical
                                        sides of D. For the condition on the top side, use a superposition
                                                                      ∞

                                                                               nπx       nπy
                                                              u(x, y) =  b n sin    sinh      .
                                                                                L         L
                                                                      n=1
                                        We need
                                                                         ∞

                                                                                  nπx       nπ K
                                                          u(x, K) = f (x) =  b n sin   sinh       .
                                                                                   L         L
                                                                         n=1
                                        This is a Fourier sine expansion of f (x) on [0, L] with coefficient b n sinh(nπ K/L). Thus, choose
                                        b n sinh(nπ K/L) to be the nth Fourier sine coefficient of f (x) on [0, L]:
                                                                           2     L       nπξ
                                                           b n sinh(nπ K/L) =    f (ξ)sin     dξ.
                                                                           L  0           L
                                        Then
                                                                     2          L        nπξ
                                                           b n =                 f (ξ)sin     dξ.
                                                                L sinh(nπ K/L)  0         L



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                                   October 14, 2010  15:27  THM/NEIL   Page-643        27410_18_ch18_p641-666
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