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618    CHAPTER 17  The Heat Equation

                                 17.2.4  Transformation of Problems
                                 As with the wave equation, it may be impossible to separate variables in a diffusion problem,
                                 depending on the partial differential equation and/or the boundary conditions. Sometimes we
                                 can transform such a problem into one to which separation of variables applies.


                         EXAMPLE 17.3
                                 A thin, homogeneous bar extends from x = 0to x = L. The left end is maintained at constant
                                 temperature T 1 and the right end at constant temperature T 2 . The initial temperature in the cross
                                 section at x is f (x).
                                    The initial-boundary value problem modeling this setting is
                                                                  2
                                                           ∂u    ∂ u
                                                              = k    for 0 < x < L,t > 0,
                                                            ∂t   ∂x 2
                                                        u(0,t) = T 1 ,u(L,t) = T 2 for t > 0,
                                 and

                                                           u(x,0) = f (x) for 0 ≤ x ≤ L.
                                 If we set u(x,t)= X(x)T (t),wemusthave X(0)T (t)= T 1 .If T 1 =0, this is satisfied by requiring
                                 that X(0) = 0. If T 1  = 0, then X(0)  = 0, and we must have T (t) = T 1 /X(0) = constant, which is
                                 not acceptable.
                                    Perturb the temperature distribution function by setting

                                                             u(x,t) = U(x,t) + ψ(x).
                                 The idea is to choose ψ to obtain a problem we know how to solve. Substitute U into the wave
                                 equation to get
                                                                      2
                                                            ∂U       ∂ U

                                                                = k     + ψ (x) .
                                                             ∂t      ∂x  2

                                 This is the standard heat equation (17.1) if ψ (x) = 0, hence, if ψ(x) = cx + d.Now
                                                           u(0,t) = T 1 = U(0,t) + ψ(0),
                                 and this is the more friendly condition U(0,t) = 0if ψ(0) = T 1 . Thus, choose d = T 1 to have
                                 ψ(x) = cx + T 1 . Next we need
                                                          u(L,t) = T 2 = U(L,t) + ψ(L),

                                 and this is just U(L,t) = 0if ψ(L) = T 2 . We need cL + T 1 = T 2 ,so c = (T 2 − T 1 )/L, and
                                                                   1
                                                            ψ(x) =  (T 2 − T 1 )x + T 1 .
                                                                   L
                                 Finally,
                                                         u(x,0) = f (x) = U(x,0) + ψ(x),

                                 so
                                                                             1
                                                  U(x,0) = f (x) − ψ(x) = f (x) −  (T 2 − T 1 )x − T 1 .
                                                                             L




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                                   October 14, 2010  15:25  THM/NEIL   Page-618        27410_17_ch17_p611-640
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