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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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