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book   Mobk070    March 22, 2007  11:7








                     24  ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
                       the time derivative term. We now have


                                                            U τ = U ξξ + 1                          (2.78)
                                                        U(0,ξ) = 0
                                                        U(τ, 0) = 0                                 (2.79)

                                                        U(τ, 1) = 0


                       2.1.20 Relocating the Nonhomogeneity
                       We have only one nonhomogeneous condition, but it’s in the wrong place. The differential
                       equation won’t separate. For example if we let U(ξ, τ) = P(ξ)G(τ) and insert this into the
                       partial differential equation and divide by PG,wefind

                                                      G (τ)    P (ξ)    1


                                                            =       +                               (2.80)
                                                        G       P      PG
                            The technique to deal with this is to relocate the nonhomogenous condition to the initial
                       condition. Assume a solution in the form U = W(ξ) + V (τ, ξ). We now have


                                                       V τ = V ξξ + W ξξ + 1                        (2.81)

                            If we set W ξξ =−1, the differential equation for V becomes homogeneous. We then set
                       both W and V equal to zero at ξ = 0and 1and V (0,ξ) =−W(ξ)


                                                                W ξξ =−1                            (2.82)
                                                       W(0) = W(1) = 0                              (2.83)

                       and


                                                            V τ = V ξξ                              (2.84)
                                                        V (0,ξ) =−W(ξ)

                                                        V (τ, 0) = 0                                (2.85)
                                                        V (τ, 1) = 0

                       The solution for W is parabolic


                                                              1
                                                         W = ξ(1 − ξ)                               (2.86)
                                                              2
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