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620 CHAPTER 17 The Heat Equation

The term F(x,t), for example, could account for a source or loss of energy within the medium.
It is routine to try separation of variables and find that it does not apply in general to this
heat equation. A different approach is needed.
As a starting point, we know that in the case that F(x,t) = 0 for all x and t the solution has
the form
∞

nπx 2 2 2
u(x,t) = c n sin e −n π kt/L .
L
n=1
Taking a cue from this case, we will attempt a solution of the current problem of the form
∞

nπx
u(x,t) = T n (t)sin . (17.4)
L
n=1
We must find the functions T n (t).If t is fixed, the left side of equation (17.4) is a function of x
and the right side is its Fourier sine expansion on [0, L], so the coefficients must be
2 L nπξ
T n (t) = u(ξ,t)sin dξ. (17.5)
L 0 L
Assume that for any t ≥ 0, F(x,t), thought of as a function of x, also can be expanded in a
Fourier sine series on [0, L]:
∞
nπξ
F(x,t) = B n (t)sin (17.6)
L
n=1
where
2 L nπξ
B n (t) = F(ξ,t)sin dξ. (17.7)
L 0 L
Differentiate equation (17.5) with respect to t to get
2 L ∂u nπξ

T (t) = (ξ,t)sin dξ. (17.8)
n
L 0 ∂t L
Substitute for ∂u/∂t from the heat equation to get
2
2k L ∂ u nπξ 2 L nπξ

T (t) = (ξ,t)sin dξ + F(ξ,t)sin dξ.
n
L 0 ∂x 2 L L 0 L
In view of equation (17.6), this equation becomes
2
2k L ∂ u nπξ

T (t) = (ξ,t)sin dξ + B n (t). (17.9)
n
L 0 ∂x 2 L
Integrate by parts twice, at the last step making use of the boundary conditions and equation
(17.5). We get
∂ u nπξ
L 2
(ξ,t)sin dξ
0 ∂x 2 L

L
∂u nπξ nπ ∂u nπξ
L
= (ξ,t)sin − (ξ,t)cos dξ
∂x L L ∂x L
0 0

L L

nπ nπξ nπ nπ nπξ
=− u(ξ,t)cos + − u(ξ,t)sin dξ
L L L L L
0 0
2
n π 2 L nπξ
=− u(ξ,t)sin dξ
L 2 0 L
2
2
2
n π L n π 2
=− T n (t) =− T n (t).
L 2 2 2L
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October 14, 2010 15:25 THM/NEIL Page-620 27410_17_ch17_p611-640