21.2 SUPERPOSITION OF SEPARATED SOLUTIONS
A bar of lengthLis initially at a temperature of 0 ◦C. One end of the bar(x=0)is held
at 0 ◦Cand the other is supplied with heat at a constant rate per unit area ofH.Findthe
temperature distribution within the bar after a timet.With our usual notation, the heat diffusion equation satisfied by the temperatureu(x, t)is
κ∂^2 u
∂x^2=
∂u
∂t,
withκ=k/(sρ), wherekis the thermal conductivity of the bar,sis its specific heat
capacity andρis its density.
The boundary conditions can be written as
u(x,0) = 0,u(0,t)=0,∂u(L, t)
∂x=
H
k,
the last of which is inhomogeneous. In general, inhomogeneous boundary conditions can
cause difficulties and it is usual to attempt a transformation of the problem into an
equivalent homogeneous one. To this end, let us assume that the solution to our problem
takes the form
u(x, t)=v(x, t)+w(x),where the functionw(x) is to be suitably determined. In terms ofvandwthe problem
becomes
κ(
∂^2 v
∂x^2+
d^2 w
dx^2)
=
∂v
∂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 choosingw(x) so as to make the new problem straightforward.
Using some physical insight, however, it is clear that ultimately (att=∞), when all
transients have died away, the endx=Lwill attain a temperatureu 0 such thatku 0 /L=H
and there will be a constant temperature gradientu(x,∞)=u 0 x/L. We therefore choose
w(x)=Hx
k.
Since the second derivative ofw(x)iszero,vsatisfies the diffusion equation and the
boundary conditions onvare now given by
v(x,0) =−Hx
k,v(0,t)=0,∂v(L, t)
∂x=0,
which are homogeneous inx.
From (21.12) a separated solution for the one-dimensional diffusion equation is
v(x, t)=(Acosλx+Bsinλx)exp(−λ^2 κt),corresponding to a separation constant−λ^2 .Ifwerestrictλto be real then all these
solutions are transient ones decaying to zero ast→∞. These are just what is required to
add tow(x) to give the correct solution ast→∞. In order to satisfyv(0,t) = 0, however,
we requireA= 0. Furthermore, since
∂v
∂x=Bexp(−λ^2 κt)λcosλx,