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book   Mobk070    March 22, 2007  11:7








                                                 THE FOURIER METHOD: SEPARATION OF VARIABLES         19
                   Solution for X(ξ) yields hyperbolic functions.


                                             X(ξ) = C cosh(λ n ξ) + D sinh(λ n ξ)               (2.40)

                   The boundary condition at ξ = 1 requires that

                                               0 = C cosh(λ n ) + D sinh(λ n )                  (2.41)

                   or, solving for C in terms of D,

                                                    C =−D tanh(λ n )                            (2.42)

                   One solution of our problem is therefore

                          U n (ξ, η) = K n sin(nπη)[sinh(anπξ/b) − cosh(anπξ/b)tanh(anπ/b)]     (2.43)


                   2.1.9  Superposition
                   According to the superposition theorem (Theorem 2) we can now form a solution as

                                   ∞

                         U(ξ, η) =    K n sin(nπη)[sinh(anπξ/b) − cosh(anπξ/b)tanh(anπ/b)]      (2.44)
                                   n=0

                   The final boundary condition (the nonhomogeneous one) can now be applied,

                                                   ∞

                                            1 =−      K n sin(nπη)tanh(anπ/b)                   (2.45)
                                                  n=1

                   2.1.10 Orthogonality
                   We have already noted that the sine function is an orthogonal function as defined on (0, 1).
                   Thus, we multiply both sides of this equation by sin(mπη)dη and integrate over (0, 1), noting
                   that according to the orthogonality theorem (Theorem 3) the integral is zero unless n = m.
                   The result is
                                      1                    1

                                                               2
                                        sin(nπη)dη =−K n     sin (nπη)dη tanh(anπ/b)            (2.46)
                                    η=0                  η=0
                                     1          n                    1
                                        [1 − (−1) ] =−K n tanh(anπ/b)                           (2.47)
                                     nπ                              2
                                                                  n
                                                        2[1 − (−1) ]
                                                K n =−                                          (2.48)
                                                       nπ tanh(anπ/b)
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