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book   Mobk070    March 22, 2007  11:7








                     22  ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
                       and

                                                   Q = C sinh(λη) + D cosh(λη)                      (2.67)

                       respectively. Remembering to apply the homogeneous boundary conditions first, we find that for
                       W(0,η) = 0, B = 0and for W(1,η) = 0, sin(λ) = 0. Thus, λ = nπ, our eigenvalues cor-
                       responding to the eigenfunctions sin(nπξ). The last homogeneous boundary condition is
                       W(ξ, 0) = 0, which requires that D = 0. There are an infinite number of solutions of the form

                                                  PQ n = K n sinh(nπη)sin(nπξ)                      (2.68)

                       2.1.16 Superposition
                       Since our problem is linear we apply superposition.

                                                        ∞

                                                  W =     K n sinh(nπη)sin(nπξ)                     (2.69)
                                                       n=1
                       Applying the final boundary condition, W(ξ, 1) = 1

                                                       ∞

                                                   1 =    K n sinh(nπ)sin(nπξ).                     (2.70)
                                                       n=1

                       2.1.17 Orthogonality
                       Multiplying both sides of Eq. (2.70) by sin(mπξ) and integrating over the interval (0, 1)

                                       1                              1
                                                       ∞

                                        sin(mπξ)dξ =      K n sinh(nπ)  sin(nπξ)sin(mπξ)dξ          (2.71)
                                                      n=0
                                      0                              0
                       The orthogonality property of the sine eigenfunction states that

                                                1

                                                                       0,   m  = n
                                                 sin(nπξ)sin(mπξ)dξ =                               (2.72)
                                                                       1/2, m = n
                                               0
                       Thus,

                                                        K n = 2/ sinh(nπ)                           (2.73)

                       and
                                                     ∞
                                                           2

                                               W =              sinh(nπη)sin(nπξ)                   (2.74)
                                                        sinh(nπ)
                                                    n=0
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