Mathematical Methods for Physics and Engineering - Matematica.NET

21.2 SUPERPOSITION OF SEPARATED SOLUTIONS◮A bar of length L is initially at a temperature of 0 ◦ C. One end of the bar (x =0)is heldat 0 ◦ C and the other is supplied with heat at a constant rate per unit area of H. Findthetemperature distribution within the bar after a time t.With our usual notation, the heat diffusion equation satisfied by the temperature u(x, t) isκ ∂2 u∂x = ∂u2 ∂t ,with κ = k/(sρ), where k is the thermal conductivity of the bar, s is its specific heatcapacity and ρ is its density.The boundary conditions can be written as∂u(L, t)u(x, 0) = 0, u(0,t)=0,= H ∂x k ,the last of which is inhomogeneous. In general, inhomogeneous boundary conditions cancause difficulties and it is usual to attempt a transformation of the problem into anequivalent homogeneous one. To this end, let us assume that the solution to our problemtakes the formu(x, t) =v(x, t)+w(x),where the function w(x) is to be suitably determined. In terms of v and w the problembecomes( )∂ 2 vκ∂x + d2 w= ∂v2 dx 2 ∂t ,v(x, 0) + w(x) =0,v(0,t)+w(0) = 0,∂v(L, t)∂x+ dw(L)dx= H k .There are several ways of choosing w(x) so as to make the new problem straightforward.Using some physical insight, however, it is clear that ultimately (at t = ∞), when alltransients have died away, the end x = L will attain a temperature u 0 such that ku 0 /L = Hand there will be a constant temperature gradient u(x, ∞) =u 0 x/L. We therefore choosew(x) = Hxk .Since the second derivative of w(x) is zero, v satisfies the diffusion equation and theboundary conditions on v are now given byv(x, 0) = − Hxk , v(0,t)=0, ∂v(L, t)=0,∂xwhich are homogeneous in x.From (21.12) a separated solution for the one-dimensional diffusion equation isv(x, t) =(A cos λx + B sin λx)exp(−λ 2 κt),corresponding to a separation constant −λ 2 .Ifwerestrictλ to be real then all thesesolutions are transient ones decaying to zero as t →∞. These are just what is required toadd to w(x) to give the correct solution as t →∞. In order to satisfy v(0,t) = 0, however,we require A = 0. Furthermore, since∂v∂x = B exp(−λ2 κt)λ cos λx,723