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628 CHAPTER 17 The Heat Equation

and

u(x,0) = f (x) for −∞ < x < ∞.

Take the Fourier transform of the heat equation with respect to x to get

2
∂u ∂ u
F = kF .
∂t ∂x 2
Because x and t are independent, F passes through ∂/∂t, and

∂u ∂

F = ˆ u(ω,t)
∂t ∂t
in which ω appears as a parameter. Next use the operational rule for the Fourier transform to
write

∂ u
2
2
F =−ω ˆu(ω,t).
∂x 2
The transformed heat equation is
∂ 2
ˆ u(ω,t) + kω ˆu(ω,t) = 0
∂t
with solutions
2
ˆ u(ω,t) = a ω e −ω kt .
Since u(x,0) = f (x),

ˆ
ˆ u(ω,0) = f (ω) = a ω .
We now have the Fourier transform of the solution:
2
ˆ u(ω,t) = f (ω)e −ω kt .
ˆ
Apply the inverse Fourier transform. Since this transform is complex valued and the solution for
a temperature distribution is real, the solution is the real part of this inverse:

2
−1 ˆ −ω kt
u(x,t) = Re F f (ω)e (t)
1 ∞ −ω kt iωx
2
ˆ
= Re f (ω)e e dω
2π −∞
1 ∞ ∞ 2
e
= Re f (ξ)e −iωξ dξ e iωx −ω kt dω
2π
−∞ −∞
1 ∞ ∞ 2
e
= Re f (ξ)e −iω(ξ−x) −ω kt dξ dω
2π
−∞ −∞
1 ∞ ∞ 2
= Re f (ξ)[cos(ω(ξ − x)) − i sin(ω(ξ − x))]e −ω kt dξ dω
2π
−∞ −∞
1 ∞ ∞ −ω kt
2
= f (ξ)cos(ω(ξ − x))e dξ dω,
2π
−∞ −∞
which is the same expression obtained by separation of variables.
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October 14, 2010 15:25 THM/NEIL Page-628 27410_17_ch17_p611-640