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The equilibrium temperature distribution is
Z h
x − h x 1
µ(x) = α + β − x < (x > − h)s(ξ) dξ,
h h κh 0
Z x Z h
x 1
µ(x) = α + (β − α) − (x − h) ξs(ξ) dξ + x (ξ − h)s(ξ) dξ .
h κh 0 x
Now we substitute u(x, t) = v(x, t) + µ(x) into Equation 9.8.

∂ ∂ 2
(v + µ(x)) = κ (v + µ(x)) + s(x)
∂t ∂x 2
00
v t = κv xx + κµ (x) + s(x)
v t = κv xx (9.9)
Since the equilibrium solution satisfies the inhomogeneous boundary conditions, v(x, t)
satisfies homogeneous boundary conditions.

v(0, t) = v(h, t) = 0.

The initial value of v is
v(x, 0) = f(x) − µ(x).
We seek a solution for v(x, t) that is a linear combination of eigen-solutions of the heat
equation. We substitute the separation of variables, v(x, t) = X(x)T(t) into Equation 9.9

T 0 X 00
= = −λ
κT X

This gives us two ordinary differential equations.
00
X + λX = 0, X(0) = X(h) = 0
0
T = −κλT.

The Sturm-Liouville problem for X(x) has the eigenvalues and orthonormal eigenfunc-
tions,
r
nπ 2 nπx
2
+
λ n = , X n = sin , n ∈ Z .
h h h
We solve for T(t).
2
T n = c e −κ(nπ/h) t .
The eigen-solutions of the partial differential equation are
r
2 nπx 2
v n (x, t) = sin e −κ(nπ/h) t .
h h

The solution for v(x, t) is a linear combination of these.
∞ r
X 2 nπx 2
v(x, t) = a n sin e −κ(nπ/h) t
h h
n=1

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