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18.5 Dirichlet Problem for Unbounded Regions 651

Solution Using the Fourier Transform
We also can solve this problem for the upper half-plane using the Fourier transform. Apply the
transform in the x variable to Laplace’s equation to obtain
2 2 2
∂ u ∂ u ∂ ˆu 2
F + F = (ω, y) − ω ˆu(ω, y) = 0.
∂x 2 ∂y 2 ∂y 2
This has the general solution
ωy
ˆ u(ω, y) = a ω e + b ω e −ωy .
ωy
We want this to be bounded. But for ω> 0, e →∞ as y →∞,so a ω = 0 for positive ω.And
e −ωy →∞ as y →∞ if ω< 0, so for negative ω,wemusthave b ω = 0. Therefore,

b ω e −ωy for ω> 0
ˆ u(ω, y) =
a ω e ωy for ω< 0.
Consolidate these cases by writing

ˆ u(ω, y) = c ω e −|ω|y .
To solve for the constants, take the transform of u(x,0) = f (x) to get

ˆ
ˆ u(ω,0) = f (ω) = c ω .
Then
ˆ
ˆ u(ω, y) = f (ω)e −|ω|y .
Finally, apply the inverse Fourier transform to get

−1 ˆ −|ω|y
u(x, y) = F f (ω)e (x)
1 ∞ −|ω|y iωx
ˆ
= f (ω)e e dω
2π −∞
1 ∞ ∞
e
= f (ξ)e −iωξ dξ e −|ω|y iωx dω
2π
−∞ −∞
1 ∞ ∞
e
= e −|ω|y −iω(ξ−x) dω f (ξ)dξ.
2π
−∞ −∞
A routine integration gives us
∞ 2y

e
e −|ω|y −iω(ξ−x) dω = .
y + (ξ − x) 2
2
−∞
Then
1 ∞ 2y
u(x, y) = f (ξ)dξ
2
2π y + (ξ − x) 2
−∞
y ∞ f (ξ)
= dξ,
π −∞ y + (ξ − x) 2
2
in agreement with the solution obtained by separation of variables.
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October 14, 2010 15:27 THM/NEIL Page-651 27410_18_ch18_p641-666