Page 647 - Advanced_Engineering_Mathematics o'neil

17.3 Solutions in an Infinite Medium 627

We must choose the coefficients so that
∞

u(x,0) = f (x) = [a ω cos(ωx) + b ω sin(ωx)]dω.
0
This is the Fourier integral of f on the real line, so the coefficients should be chosen as the
Fourier integral coefficients of f :
1 ∞ 1 ∞
a ω = f (ξ)cos(ωξ)dξ and b ω = f (ξ)sin(ωξ)dξ.
π π
−∞ −∞
EXAMPLE 17.7
Suppose the initial temperature function is f (x) = e −|x| . Compute the coefficients
1 ∞ 2 1
a ω = e −|ξ| cos(ωξ)dξ =
π −∞ π 1 + ω 2
and b ω = 0, because e −|x| sin(ωx) is an odd function. The solution is
2 ∞ 1 −ω kt
2
u(x,t) = cos(ωx)e dω.
π 0 1 + ω 2
In general, if we put the integrals for a ω and b ω into the solution (17.11) and argue as we did
for the solution of the wave equation on the line (equation (16.8)), we obtain an alternative form
of this solution:
1 ∞ ∞ 2
u(x,t) = cos(ω(x − ξ)) f (ξ)e −ω kt dω dξ. (17.12)
π 0 −∞
It is possible to rewrite this solution in terms of a single integral. To do this, first write the solution
(17.12) as
1 ∞ ∞ −ω kt
2
u(x,t) = cos(ω(x − ξ)) f (ξ)e dω dξ.
2π
−∞ −∞
Now we need the following integral. If α and β are real numbers with β = 0,
∞ 2 αζ √ 2 2

e −ζ cos dζ = πe −α /4β . (17.13)
β
−∞
A derivation of this integral is sketched in Problem 11. In this integral, let
√ √
ξ = ktω, α = x − ξ, and β = kt
to obtain
√
∞ 2 π 2

e −ω kt cos(ω(x − ξ))dω = √ e −(x−ξ) /4kt .
kt
−∞
Now the solution (17.12) can be written
1 ∞ 2
u(x,t) = √ f (ξ)e −(x−ξ) /4kt dξ. (17.14)
2 πkt −∞
17.3.2 Solution by Fourier Transform

We will illustrate the use of the Fourier transform to solve the problem
2
∂u ∂ u
= k for −∞ < x < ∞,t > 0
∂t ∂x 2

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October 14, 2010 15:25 THM/NEIL Page-627 27410_17_ch17_p611-640