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book   Mobk070    March 22, 2007  11:7








                                                 THE FOURIER METHOD: SEPARATION OF VARIABLES         23
                   Recall that


                                           V = W(ξ, 1 − η)   and U = V + W




                   2.1.18 Lessons
                   If there are two nonhomogeneous boundary conditions break the problem into two problems
                   that can be added (since the equations are linear) to give the complete solution. If you are
                   unsure of the sign of the separation constant just assume a sign and move on. Listen to what the
                   mathematics is telling you. It will always tell you if you choose wrong.

                   Example 2.4. A Non-homogeneous Heat Conduction Problem

                   Consider now the arrangement above, but with a heat source, and with both boundaries held
                   at the initial temperature u 0 . The heat source is initially zero and is turned on at t = 0 . The
                                                                                                +
                   exercise illustrates the method of solving the problem when the single nonhomogeneous condition is in
                   the partial differential equation rather than one of the boundary conditions.



                                                     ρcu t = ku xx + q                          (2.75)

                                                    u(0, x) = u 0
                                                    u(t, 0) = u 0                               (2.76)
                                                    u(t, L) = u 0



                   2.1.19 Scales and Dimensionless Variables
                   Observe that the length scale is still L, so we define ξ = x/L. Recall that k/ρc = α is the
                   diffusivity. How shall we nondimensionalize temperature? We want as many ones and zeros
                   in coefficients in the partial differential equation and the boundary conditions as possible.
                   Define U = (u − u 0 )/S, where S stands for “something with dimensions of temperature” that
                   we must find. Dividing both sides of the partial differential equation by q and substituting
                   for x


                                                   2
                                                  L SρcU t   kSU ξξ
                                                           =        + 1                         (2.77)
                                                     q         q

                   Letting S = q/k leads to one as the coefficient of the first term on the right-hand side.
                                                                       2
                   Choosing the same dimensionless time as before, τ = αt/L results in one as the coefficient of
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