Mathematical Methods for Physics and Engineering - Matematica.NET

21.4 INTEGRAL TRANSFORM METHODS◮An infinite metal bar has an initial temperature distribution f(x) along its length. Findthe temperature distribution at a later time t.We are interested in values of x from −∞ to ∞, which suggests Fourier transformationwith respect to x. Assuming that the solution obeys the boundary conditions u(x, t) → 0and ∂u/∂x → 0as|x| →∞, we may Fourier-transform the one-dimensional diffusionequation (21.71) to obtain∫κ ∞√2π−∞∂ 2 u(x, t)∂x 2 exp(−ikx) dx = 1 √2π∂∂t∫ ∞−∞u(x, t)exp(−ikx) dx,where on the RHS we have taken the partial derivative with respect to t outside theintegral. Denoting the Fourier transform of u(x, t) byũ(k, t), and using equation (13.28) torewrite the Fourier transform of the second derivative on the LHS, we then have−κk 2 ∂ũ(k, t)ũ(k, t) = .∂tThis first-order equation has the simple solutionũ(k, t) =ũ(k, 0) exp(−κk 2 t),where the initial conditions giveũ(k, 0) = √ 1 ∫ ∞u(x, 0) exp(−ikx) dx2π −∞= √ 1 ∫ ∞f(x)exp(−ikx) dx = ˜f(k).2πThus we may write the Fourier transform of the solution as−∞ũ(k, t) =˜f(k)exp(−κk 2 t)= √ 2π ˜f(k)˜G(k, t), (21.73)where we have defined the function ˜G(k, t) =( √ 2π) −1 exp(−κk 2 t). Since ũ(k, t) canbewritten as the product of two Fourier transforms, we can use the convolution theorem,subsection 13.1.7, to write the solution asu(x, t) =∫ ∞−∞G(x − x ′ ,t)f(x ′ ) dx ′ ,where G(x, t) is the Green’s function for this problem (see subsection 15.2.5). This functionis the inverse Fourier transform of ˜G(k, t) and is thus given byG(x, t) = 12π= 12π∫ ∞−∞∫ ∞−∞Completing the square in the integrand we findG(x, t) = 1 ( ) ∫ ∞2π exp − x24κt= 1 (2π exp − x24κt= 1 √4πκtexpexp(−κk 2 t)exp(ikx) dk[exp −κt) ∫ ∞(− x24κtexp−∞exp−∞),(k 2 − ixκt k )]dk.[−κt(k − ix ) ] 2dk2κt(−κtk ′2) dk ′where in the second line we have made the substitution k ′ = k − ix/(2κt), and in the last749