Page 23 - Essentials of applied mathematics for scientists and engineers
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book   Mobk070    March 22, 2007  11:7








                                                 THE FOURIER METHOD: SEPARATION OF VARIABLES         13
                        We first note that the problem has a fundamental length scale, so that if we define another
                   space variable ξ = x/L, the partial differential equation can be written as

                                                    −2
                                            ρcu t = L ku ξξ  0 <ξ < 1   t < 0                    (2.3)
                                                                                     2
                   Next we note that if we define a dimensionless time-like variable as τ = αt/L , where α = k/ρc
                   is called the thermal diffusivity,wefind


                                                        u τ = u ξξ                               (2.4)
                   We now proceed to nondimensionalize and normalize the dependent variable and the boundary
                   conditions. We define a new variable

                                                 U = (u − u 1 )/(u 0 − u 1 )                     (2.5)

                   Note that this variable is always between 0 and 1 and is dimensionless. Our boundary value
                   problem is now devoid of constants.


                                                       U τ = U ξξ                                (2.6)
                                                       U(τ, 0) = 0
                                                       U ξ (τ, 1) = 0                            (2.7)
                                                       U(0,ξ) = 1

                   All but one of the boundary conditions are homogeneous. This will prove necessary in our analysis.


                   2.1.2  Separation of Variables
                   Begin by assuming U =  (τ) (ξ). Insert this into the differential equation and obtain

                                                  (ξ)  τ (τ) =  (τ)  ξξ (ξ).                     (2.8)

                   Next divide both sides by U =   ,

                                                      τ     ξξ     2
                                                       =      =±λ                                (2.9)

                   The left-hand side of the above equation is a function of τ only while the right-hand side is a
                   function only of ξ. This can only be true if both are constants since they are equal to each other.
                    2
                   λ is always positive, but we must decide whether to use the plus sign or the minus sign. We
                   have two ordinary differential equations instead of one partial differential equation. Solution
                                                         2
                                                                     2
                   for   gives a constant times either exp(−λ τ)orexp(+λ τ). Sinceweknowthat U is always
                   between 0 and 1, we see immediately that we must choose the minus sign. The second ordinary
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