Page 603 - Advanced_Engineering_Mathematics o'neil

16.3 Wave Motion in an Infinite Medium 583

Write this differential equation as
∂ 2 2 2
ˆ y(ω,t) + c ω ˆy(x,t) = 0.
∂t 2
Think of this as an ordinary differential equation for ˆy(ω,t) with t as the variable and ω carried
along as a parameter. The general solution is
ˆ y(ω,t) = a ω cos(ωct) + b ω sin(ωct).
To solve for the coefficients to satisfy the initial conditions, transform the initial data. First,
ˆ y(ω,0) = a ω = F[y(x,0)](ω) = F[ f (x)](ω) = f (ω),
ˆ
the transform of the initial position function. Next,
∂ ˆy ∂y
(ω,0) = cωb ω = F (x,0) (ω,0)
∂t ∂t
= F[g(x)](ω) =ˆg(ω),
which is the transform of the initial velocity function. Therefore,
1
b ω = ˆ g(ω).
ωc
Now
1
ˆ
ˆ y(ω,t) = f (ω)cos(ωct) + ˆ g(ω)sin(ωct).
ωc
This is the Fourier transform of the solution. Invert this to find the solution
1 ∞ 1 iωx
ˆ
y(x,t) = f (ω)cos(ωct) + ˆ g(ω)sin(ωct) e dω.
2π −∞ ωc
In the case that the string is released from rest, g(x) = 0 and this solution is
1 ∞
ˆ
y(x,t) = f (ω)cos(ωct)e iωx dω. (16.11)
2π
−∞
We claim that this solution (16.11) obtained by using the Fourier transform agrees with the
solution (16.8) obtained by separation of variables and the Fourier integral. For the moment,
denote the solution (16.11) by Fourier transform as y tr (x,t). Manipulate y tr (x,t) as follows:
1 ∞ iωx
ˆ
y tr (x,t) = f (ω)cos(ωct)e dω
2π
−∞
1 ∞ ∞
= f (ξ)e −iωξ dξ cos(ωct)e iωx dω
2π
−∞ −∞
1 ∞ ∞
= e −iω(ξ−x) cos(ωct) f (ξ)dω dξ
2π
−∞ −∞
1 ∞ ∞
= [cos(ω(ξ − x)) − i sin(ω(ξ − x))]cos(ωct) f (ξ)dω dξ.
2π
−∞ −∞
Since y(x,t) must be real-valued, the solution is actually the real part of this integral. Therefore,
1 ∞ ∞
y tr (x,t) = cos(ω(ξ − x))cos(ωct) f (ξ)dω dξ.
2π
−∞ −∞
Finally, the integrand is an even function of ω,so
1 ∞ ∞
y tr (x,t) = cos(ω(ξ − x))cos(ωct) f (ξ)dω dξ,
π
−∞ 0
and this is the solution (16.8).

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October 14, 2010 15:23 THM/NEIL Page-583 27410_16_ch16_p563-610